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a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
=>4x-27=1
hay x=7
b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)
\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)
=>39x+6=15
hay x=3/13
c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
hay x=14
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a: =>x^2+4x-4x+1=0
=>x^2+1=0
=>Loại
b: =>2x-6+4=2x+2
=>-2=2(loại)
c: =>2(x+3)-2x-1=1
=>6-1=1
=>5=1(loại)
d =>x+3=0
=>x=-3(loại)
e: =>x^2-3x^2+3x-3x-2=0
=>-2x^2-2=0
=>x^2+1=0
=>Loại
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
\(a.\left(2-3x\right)\left(x^2+2x+3\right)=0.\)
\(\left(2-3x\right)=0\)
\(\left(x^2+2x+3\right)=0\)
\(TH1:2-3x=0\Leftrightarrow x=\frac{-2}{-3}\)
\(TH2:x^2+2x+3=0\Leftrightarrow\left(x^2+2x+1\right)+3\Leftrightarrow\left(x+1\right)^2+3>0\)
b) \(3x-3x=5+2\) ( vô nghiệm)
c) vô nghiệm
d-\(x^2-5x-6=0\Leftrightarrow\left(x^2-x\right)+\left(6x-6\right)\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
vậy ...
x=1
x=-6
E) \(\frac{2\left(x-3\right)^2}{3}=\frac{3x^2}{2}\) quy đồng khử mẫu ta được
\(4\left(x-3\right)^2-9x^2=0\Leftrightarrow4\left(x-3\right)^2-\frac{4.1.9x^2}{4}\) rút 4 ta được
\(4\left\{\left(x-3\right)^2-\frac{9x^2}{4}\right\}=0\Leftrightarrow4\left\{\left(x-3\right)^2-\left(\frac{3}{2}x\right)^2\right\}\Leftrightarrow4\left(x-3+\frac{3}{2}x\right)\left(x-3-\frac{3}{2}x\right)=0\) ( hằng đẳng thức số 3 )
tích = 0
vậy ....
F) trị tuyệt đối + bình phương của 1 số thực luôn lớn hơn hoặc = 0( định lí Pain)
phá trị tuyệt đối ta được
\(\left(x+5\right)^2-\left(3x-2\right)^2=0\)
\(\left(x+5-3x-2\right)\left(x+5+3x-2\right)=0\) ( hẳng đẳng thức số 3 )
tích = 0 suy ra 2 TH vậy .....
g) câu G bạn lên coccoc math bạn ghi là nó ra kết quả phân tích thành nhân tử chứ làm = tay vừa dài vừa hại não :)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24=0\)
\(x\left(x-5\right)x\left(x^2-5x+10\right)=0\) ( coccoc math)
\(\left(x^2-5x+10\right)=0\Leftrightarrow\left(x^2-\frac{2x.5}{2}+\left(\frac{5}{2}\right)^2\right)+10-\frac{25}{4}=0\) ( 10-25/4) = 15/4
\(\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\) ( vô nghiệm)
vậy....
x3 + 3x2 + 3x = 7
<=> x3 + 3x2 + 3x - 7 = 0
<=> (x - 1)(x2 + 4x + 7) = 0
<=> x - 1 = 0 hoặc x2 + 4x + 7 khác 0
<=> x - 1 = 0
<=> x = 1
a) ( x2 + 3 x + 2 ) . ( x2 + 3x+ 3 ) - 2 =0
<=>x4 + 3x3 + 3x2 + 3x3 + 9x2 + 9x + 2x2 + 6x + 6 - 2 = 0
<=> x4 + 6x3 + 14x2 + 15x + 4 = 0
<=> x4 + 3x3 + 3x3 + x2 + 9x2 + 4x2 + 3x + 12x + 4 = 0
<=> x2 . ( x2 +3x + 1 ) + 3x . ( x2 +3x + 1 ) + 4. ( x2 + 3x + 1 ) = 0
<=> ( x2 + 3x + 1 ) . ( x2 + 3x + 4 ) = 0
<=> \(\orbr{\begin{cases}x^2+3x+1=0\\x^2+3x+4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)
\(x\notinℝ\)
<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)
Nghiệm cuối cùng là : x1 = \(\frac{-3+\sqrt{5}}{2}\);x2 = \(\frac{-3-\sqrt{5}}{2}\)
b) ( x + 1 ) . ( x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0
<=> ( x2 + 2x + x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0
<=> ( x2 + 3.x + 2 ) . ( x+3) . ( x + 4 ) -24 = 0
<=> ( x3 + 3.x 2 + 3.x2 + 9x + 2x + 6 ) . ( x + 4 ) - 24 = 0
<=> ( x3 + 3x + 2 ) . ( x + 3 ) .( x + 4 ) = 0
<=> ( x3 + 3x2 + 3x2 + 9x + 2x + 6 ) . ( x + 4) - 24 = 0
<=> ( x3 + 6.x2 + 11.x + 6 ) . ( x + 4 ) -24 = 0
<=> x4 + 4.x3 + 6.x3 + 24.x2 + 11.x2 + 44.x + 6.x + 24 - 24 =0
<=> x4 + 10.x3+ 35. x2 + 50.x = 0
<=> x. ( x3 + 10.x2 + 35 .x + 50 ) = 0
<=> x. ( x3 + 5.x2 +5.x2 + 25.x+ 10 + 50 ) = 0
<=> x. [ x2 . ( x+5 ) + 5.x. ( x+5 ) + 10.( x + 5 ) ] = 0
<=> x. ( x + 5 ) . ( x2 + 5.x + 10 ) = 0
=> \(\hept{\begin{cases}x=0\\x+5=0\\x^2+5.x+10=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\x=-5\\x\notinℝ\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
Nghiệm cuối cùng là : x1 = -5 ; x2 = 0
c) x3 + 3.x2 + 3x = 7
<=> x3 + 3.x2 + 3x - 7 = 0
<=> ( x + 1 )3 - 8 = 0
<=> ( x + 1 )3 = 8
<=> ( x + 1 ) 3 = 23
<=> x + 1 = 2
<=> x =1
Vậy x = 1