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22 tháng 5 2022
a: \(\left(2x-3\right)\left(3x^2+1\right)-6x\left(x^2-x+1\right)+3x^2-2x=10\)
\(\Leftrightarrow6x^3+2x-9x^2-3-6x^3+6x^2-6x+3x^2-2x=10\)
\(\Leftrightarrow-6x-3=10\)
=>-6x=13
hay x=-13/6
b: \(\Leftrightarrow3x^2-3x+x-2-3x^2+5x=-8-5x\)
=>3x-2=-5x-8
=>8x=-6
hay x=-3/4
c: \(\Leftrightarrow64x^3-27-64x^3+32x^2-32x^2+x=20\)
=>x-27=20
hay x=47
KT
0
PN
22 tháng 5 2021
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
PN
22 tháng 5 2021
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
TT
23 tháng 4 2019
a. \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left[3\left(2x-1\right)-5\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
b. \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow-\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)
c. \(\left(2x+1\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow3x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d. \(2x^3+3x^2-32x=48\)
\(\Leftrightarrow2x^3+3x^2-32x-48=0\)
\(\Leftrightarrow\left(2x^3-8x^2\right)+\left(5x^2-20x\right)-\left(12x-48\right)=0\)
\(\Leftrightarrow2x^2\left(x-4\right)+5x\left(x-4\right)-12\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x^2+5x-12\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\frac{3}{2}\end{matrix}\right.\)
e. \(x^2+2x-15=0\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(5x-15\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
NT
1
28 tháng 2 2022
a: \(x^2\left(2x-3\right)+8x-12=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x^2+4\right)=0\)
=>2x-3=0
hay x=3/2
b: \(\Leftrightarrow\left(2x-5\right)\left(2x+10\right)-\left(2x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+10-x+1\right)=0\)
=>(2x-5)(x+11)=0
=>x=5/2 hoặc x=-11
c: \(\Leftrightarrow2x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{0;4;-4\right\}\)
KU
1
18 tháng 4 2020
a)3(x-1)(2x-1)-5(x+8)(x-1)=0
<=>(x-1)(6x-3-5x-40)=0
<=>(x-1)(x-43)=0
b)2x^3+3x^2-32x-48=0
<=>x^2(2x+3)-16(2x+3)=0
<=>(2x+3)(x-4)(x+4)=0
học tốt