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a/ \(x^2\left(x-5\right)+5-x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
Vậy...
b/ \(3x^4-9x^3=-9x^2+27x\)
\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)
\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x^3+9x\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)
Vì \(x^2+3>0\forall x\)
\(\Leftrightarrow3x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy..
c/ \(x^2\left(x+8\right)+x^2=-8x\)
\(\Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\)
\(\Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\)
\(\Leftrightarrow x\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\\x=-1\end{matrix}\right.\)
Vậy...
d/ \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x-2\right)^2+1\right]=0\)
Vì \(\left(x-2\right)^2+1>0\forall x\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy..
Úi, câu d bạn nên làm theo cách của bạn trên đúng hơn nha :< Mình nghĩ câu d mình lập luận bị sai rồi ó
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình trên là \(-2\)
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
Bài 1:
a)-x^2+4x-5
=-(x2-4x+5)<0 với mọi x
=>-x^2+4x-5<0 với mọi x
b)x^4+3x^2+3
\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x
=>x^4+3x^2+3>0 với mọi x
c) bn xét từng th ra
Bài 2:
a)9x^2-6x-3=0
=>3(3x2-2x-1)=0
=>3x2-2x-1=0
=>3x2+x-3x-1=0
=>x(3x+1)-(3x+1)=0
=>(x-1)(3x+1)=0
b)x^3+9x^2+27x+19=0
=>(x+1)(x2+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)
- Với x+1=0 =>x=-1
- Với x2+8x+19 =>vô nghiệm
c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
=>x3-25x-x3-8=3
=>-25x-8=3
=>-25x=1
=>x=-11/25
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
Bài 1: Tìm x
a) Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-9=0\)
\(\Leftrightarrow-12x-24=0\)
\(\Leftrightarrow-12x=24\)
hay x=-2
Vậy: x=-2
b) Ta có: \(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)-11\left(x-1\right)\left(x+1\right)-6=0\)
\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11\left(x^2-1\right)-6=0\)
\(\Leftrightarrow11x^2+6x+12-11x^2+11=0\)
\(\Leftrightarrow6x+23=0\)
\(\Leftrightarrow6x=-23\)
hay \(x=-\frac{23}{6}\)
Vậy: \(x=-\frac{23}{6}\)
c) Ta có: \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
hay \(x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
d) Ta có: \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Vậy: x=-3
a) (2x + 1)2 - 4(x + 2)2 = 9
4x2 + 4x + 1 - 4(x2 + 4x + 4) = 9
4x2 + 4x + 1 - 4x2 - 16x - 16 = 9
-12x - 15 = 9
-12x = 9 + 15
-12x = 24
x = 12 : (-2)
x = -2
b) (3x - 1)2 + 2(x + 3)2 + 11(x + 1)(1 - x) = 6
9x2 - 6x + 1 + 2(x2 + 6x + 9) - 11(x + 1)(x - 1) = 6
9x2 - 6x + 1 + 2x2 + 12x + 18 - 11(x2 - 1) = 6
9x2 - 6x + 1 + 2x2 + 12x + 18 - 11x2 + 11 = 6
6x + 30 = 6
6x = 6 - 30
6x = -24
x = -24 : 6
x = -4
c) 8x3 - 12x2 + 6x - 1 = 0
(2x)3 - 3.(2x)2.1 + 3.2x.12 - 13 = 0
(2x - 1)3 = 0
2x - 1 = 0
2x = 1
x = 1/2
d) x3 + 9x2 + 27x + 27 = 0
x3 + 3.x2.3 + 3.x.32 + 33 = 0
(x + 3)3 = 0
x + 3 = 0
x = 0 - 3
x = -3
a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)
b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x\left(x-4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;4\right\}\)
c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)
\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)
d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)
e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)
g. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)
a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
b: =(1-2x)(1+2x)
c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d: =(x+3)^3
e: \(=\left(2x-y\right)^3\)
f: =(x+2y)(x^2-2xy+4y^2)
Bài 2:
a: \(x^2\left(x-2\right)+2-x=0\)
=>\(x^2\left(x-2\right)-\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x^2-1\right)=0\)
=>(x-2)(x+1)(x-1)=0
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=-1\\x=1\end{matrix}\right.\)
b: \(x^2-9x^3=x^2-9x\)
=>\(9x^3=9x\)
=>\(x^3=x\)
=>\(x^3-x=0\)
=>\(x\cdot\left(x^2-1\right)=0\)
=>x(x-1)(x+1)=0
=>\(\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
c: \(x\left(x+2\right)+x^2=-2x\)
=>\(x\left(x+2\right)+x^2+2x=0\)
=>2x(x+2)=0
=>x(x+2)=0
=>\(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d: \(\left(x+1\right)\left(x^2+4\right)=x^2+x\)
=>\(\left(x+1\right)\left(x^2+4\right)-x\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^2-x+4\right)=0\)
mà \(x^2-x+4=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>=\dfrac{15}{4}\forall x\)
nên x+1=0
=>x=-1
sao phần a câu 1 lại có +-x hả cậu