a,( 3x+1)2-(x-2)2=0
=>(3x+1-x+2)(3x+1+x-2)=0
=>(2x+3)(4x-1)=0
=>vậy x=\(\frac{-3}{2}hoặcx=\frac{1}{4}\)
b.x3-x=x2+x
=>x(x2-1)=x(x+1)
=>x2-1=x+1
=>x2-x-2=0
=>x=2 hoặc x=-1
c.(x+2)2=9(x-2)2
=>x+2-3x+6=0 hoặc x+2+3x+6=0
=>x=4 hoặc x=-2
d, tương tự nhé
a) Ta có: \(\left(3x+1\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(3x+1+x-2\right)\left(3x+1-x+2\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=1\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{-3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{4};\frac{-3}{2}\right\}\)
b) Ta có: \(x^3-x=x^2+x\)
\(\Leftrightarrow x^3-x-x^2-x=0\)
\(\Leftrightarrow x^3-2x-x^2=0\)
\(\Leftrightarrow x^3-x^2-2x=0\)
\(\Leftrightarrow x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;2;-1}
c) Ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2=9\left(x-2\right)^2\)
\(\Leftrightarrow\left(x+2\right)^2=\left[3\left(x-2\right)\right]^2\)
\(\Leftrightarrow\left(x+2\right)^2=\left(3x-6\right)^2\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+6\right)\left(x+2+3x-6\right)=0\)
\(\Leftrightarrow\left(-2x+8\right)\left(4x-4\right)=0\)
\(\Leftrightarrow-2\left(x-4\right)\cdot4\cdot\left(x-1\right)=0\)
Vì -2≠0 và 4≠0
nên \(\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy: x∈{1;4}
d) Ta có: \(\left(x-1\right)^2-1+x^2=\left(3x+2\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=6x^2+3x+4x+2\)
\(\Leftrightarrow2x^2-2x=6x^2+7x+2\)
\(\Leftrightarrow2x^2-2x-6x^2-7x-2=0\)
\(\Leftrightarrow-4x^2-9x-2=0\)
\(\Leftrightarrow-4x^2-8x-x-2=0\)
\(\Leftrightarrow-4x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{-1}{4}\right\}\)
a) (3x + 1)2 - (x - 2)2 = 0
⇔ (3x + 1 - x + 2)(3x + 1 + x - 2) = 0
⇔ (2x + 3)(4x - 1) = 0
⇔ \(\left[{}\begin{matrix}2x+3=0\\4x-1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy nghiệm của pt là x = \(\frac{-3}{2}\); x = \(\frac{1}{4}\)
b) x3 - x = x2 + x
⇔ x3 - x2 - 2x = 0
⇔ x(x2 - x - 2) = 0
⇔ x(x - 2)(x + 1) = 0
⇔ \(\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)
Vậy nghiệm của pt là x = 0; x = 2; x = -1
c) (x + 2)2 = 9(x2 - 4x + 4)
⇔ (x + 2)2 - [3(x - 2)]2 = 0
⇔ (x + 2 - 3x + 6)(x + 2 + 3x - 6) = 0
⇔ (-2x + 8)(4x - 4) = 0
⇔ \(\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là x = 4; x = 1
d) (x - 1)2 - 1 + x2 = (3x + 2)(2x + 1)
⇔ x2 - 2x + 1 - 1 + x2 = 6x2 + 4x + 3x + 2
⇔ 4x2 + 9x + 2 = 0
⇔ 4x2 + x + 8x + 2 = 0
⇔ x(4x + 1) + 2(4x + 1) = 0
⇔ (4x + 1)(x + 2) = 0
⇔ \(\left[{}\begin{matrix}x=\frac{-1}{4}\\x=-2\end{matrix}\right.\)
Vậy nghiệm của pt là x = -2; x = \(\frac{-1}{4}\)
