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Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
a) Ta có: 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)
b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
hay \(x=\frac{15}{8}\)
Vậy: \(x=\frac{15}{8}\)
c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)
\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)
\(\Leftrightarrow-6x^2-11x=0\)
\(\Leftrightarrow6x^2+11x=0\)
\(\Leftrightarrow x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)
d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
\(\Leftrightarrow14x^2=-18\)
mà \(14x^2\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
b)
\(2x\cdot\left(2x-3\right)=\left(3-2x\right)\cdot\left(2-5x\right)\\ \Leftrightarrow-2x\cdot\left(3-2x\right)-\left(3-2x\right)\cdot\left(2-5x\right)=0\\ \Leftrightarrow\left(3-2x\right)\cdot\left(-2x-2+5x\right)=0\\ \Leftrightarrow\left(3-2x\right)\cdot\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3-2x=0\\3x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
c)
\(2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^3+6x^2-x^2-3x=0\\ \Leftrightarrow x\cdot\left(2x^2+6x-x-3\right)=0\\ \Leftrightarrow x\cdot\left(-3+6x-x+2x^2\right)=0\\ \Leftrightarrow x\cdot\left[-3\cdot\left(1-2x\right)-x\cdot\left(1-2x\right)\right]=0\\ \Leftrightarrow x\cdot\left(-3-x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\-3-x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
d)
\(x^2-5x+6=0\\ \Leftrightarrow x^2-3x-2x+6=0\\ \Leftrightarrow6-2x-3x+x^2=0\\ \Leftrightarrow2\cdot\left(3-x\right)-x\cdot\left(3-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
e)
\(\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5+x+2\right)\cdot\left(2x+5-x-2\right)=0\\ \Leftrightarrow\left(3x+7\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+7=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{7}{3}\\x=-3\end{matrix}\right.\)
a) \(\left(x+3\right)\left(x+5\right)+\left(x+3\right)\left(3x-4\right)=0\)
➜\(\left(x+3\right)\left(x+5+1+3x-4\right)=0\)
➜\(\left[{}\begin{matrix}x+3=0\\x+3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
Mk đang hok zoom sorry nha!!!
a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)-3=-3\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3-3=-3\)
\(\Leftrightarrow14x=0\)
\(\Leftrightarrow x=0\)
Vậy pt có nghiệm duy nhất x = 0.
b) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=7\)
\(\Leftrightarrow18x-2=7\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy pt có nghiệm duy nhất \(x=\frac{1}{2}\)
c) \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow36x^2-24x+4+25x^2-20x+4-60x^2+33x-8=0\)
\(\Leftrightarrow x^2-11x=0\)
\(\Leftrightarrow x\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;11\right\}\)
d) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-4x+32=1\)
\(\Leftrightarrow41-10x=1\)
\(\Leftrightarrow-10x=40\)
\(\Leftrightarrow x=-4\)
Vậy pt có nghiệm duy nhất x = -4.
e) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7x^2+36=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2=0\)
\(\Leftrightarrow8x=-13\)
\(\Leftrightarrow x=-\frac{13}{8}\)
Vậy pt có nghiệm duy nhất \(x=-\frac{13}{8}\)
Bài 1:
a) 5(x-3)-4=2(x-1)
\(\Leftrightarrow5x-15-4=2x-2\)
\(\Leftrightarrow5x-19-2x+2=0\)
\(\Leftrightarrow3x-17=0\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)
Vậy: \(x=\frac{17}{3}\)
b) 5-(6-x)=4(3-2x)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow-1+x-12+8x=0\)
\(\Leftrightarrow-13+9x=0\)
\(\Leftrightarrow9x=13\)
\(\Leftrightarrow x=\frac{13}{9}\)
Vậy: \(x=\frac{13}{9}\)
c) (3x+5)(2x+1)=(6x-2)(x-3)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
\(\Leftrightarrow x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8\)
\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)
\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow x=1\)
Vậy:x=1
Bài 2:
a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)
\(\Leftrightarrow4x-10x-15x-3x+60=0\)
\(\Leftrightarrow-24x+60=0\)
\(\Leftrightarrow-24x=-60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy: \(x=\frac{5}{2}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)
\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)
\(\Leftrightarrow-3x=0\)
\(\Leftrightarrow x=0\)
Vậy: x=0
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow3x=-48\)
\(\Leftrightarrow x=-16\)
Vậy: x=-16
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)
\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)
\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)
\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow-13x=-143\)
\(\Leftrightarrow x=11\)
Vậy: x=11
e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)
\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)
\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)
\(\Leftrightarrow45x-18-24-28x+60x-420=0\)
\(\Leftrightarrow77x-462=0\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
Vậy:x=6
Bài 3:
a) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)
Vì \(2\ne0\)
nên \(\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)
b) \(\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)
c) \(\left(2x+1\right)\left(x^2+2\right)=0\)
Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy: \(x=\frac{-1}{2}\)
d) \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0\)
Ta có: \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta lại có \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)
Ta có: \(4\ne0\)(4)
Từ (3) và (4) suy ra
2x-1=0
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
Bài 4:
a) \(\left(x-2\right)\left(2x+3\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2\)
\(\Leftrightarrow2x^2-x-6=x^2-3x+2\)
\(\Leftrightarrow2x^2-x-6-x^2+3x-2=0\)
\(\Leftrightarrow x^2+2x-8=0\)
\(\Leftrightarrow x^2+2x+1-9=0\)
\(\Leftrightarrow\left(x+1\right)^2-3^2=0\)
\(\Leftrightarrow\left(x+1-3\right)\left(x+1+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
b) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
c) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)
\(\Leftrightarrow-8x^2+40x-32=0\)
\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)
\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)
Vì \(-8\ne0\)
nên \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{1;4\right\}\)
e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)
\(\Leftrightarrow7x^2+58x+115=0\)
\(\Leftrightarrow7x^2+23x+35x+115=0\)
\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)
\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)
Bài 5:
a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\x=-1\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)
b) \(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow2x^2-2x+x^2+2x-3=0\)
\(\Leftrightarrow3x^2-3=0\)
\(\Leftrightarrow3\left(x^2-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-1\right\}\)
c) \(x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)
Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)
Từ (5) và (6) suy ra
\(\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy: x=-1
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
P/S : Tại vì đề bài không cho biết tìm x hay giải phương trình nên mình sẽ giải phương trình luôn nhé
(x - 5)(3 + 5x)(6x + 1) = 0 <=> \(\hept{\begin{cases}x-5=0\\3+5x=0\\6x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=5\\x=-\frac{3}{5}\\x=-\frac{1}{6}\end{cases}}\)
Vậy \(S=\left\{5;-\frac{3}{5};-\frac{1}{6}\right\}\)
(x - 3)(3x + 5) = (2x - 6)(x + 1)
<=> 3x2 + 5x - 9x - 15 = 2x2 + 2x - 6x - 6
<=> 3x2 + 5x - 9x - 15 - 2x2 - 2x + 6x + 6 = 0
<=> x2 - 9 = 0
<=> (x - 3)(x + 3) = 0
<=> \(\hept{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy \(S=\left\{3;-3\right\}\)
(2x2 + 1)(x2 - 2) = 0
Vì 2x2 + 1 > 0 nên không tìm được x
x2 - 2 = 0 <=> \(x^2-\left(\sqrt{2}\right)^2=0\)
<=> \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
<=> \(\hept{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
Vậy \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)
a, (x – 5) (3 + 5x) (6x + 1) = 0
\(\Rightarrow\)Hoặc x - 5 = 0 hoặc 3 + 5x = 0 hoặc 6x + 1 = 0
\(\Leftrightarrow\)Hoặc x = 5 hoặc x = -3/5 hoặc x = -1/6
b, (x – 3)(3x + 5) = (2x - 6)(x + 1)
\(\Leftrightarrow\)(x – 3) (3x + 5) - (2x - 6) (x + 1) = 0
\(\Leftrightarrow\)\(3x^2+5x-9x-15-2x^2-2x+6x+6=0\)
\(\Leftrightarrow x^2-9=0\)
\(\Leftrightarrow\)( x - 3 ) ( x + 3) = 0
\(\Leftrightarrow\)Hoặc x = 3 hoặc x = -3
c, (2x2 + 1)(x2 - 2) = 0
\(\Leftrightarrow\)Hoặc \(2x^2+1=0\)hoặc \(x^2-2=0\)
\(\Leftrightarrow\)Hoặc \(2x^2=-1\)(vô lí) hoặc \(x^2=2\)
\(\Leftrightarrow\)\(x=\sqrt{2}\) hoặc \(x=-\sqrt{2}\)
\(\Leftrightarrow\)