Question

Tìm x

a, x\(^2\)-x(x+2)-1=2

b, x\(^2\)=(2x-1)\(^2\)

Answer 1

\(x^2-x^2-2x-1-2=0\)

\(-2x-3=0\Leftrightarrow x=\dfrac{-2}{3}\)

\(\left(x-2x+1\right)\left(x+2x-1\right)=0\)

\(\left[{}\begin{matrix}-x+1=0\\3x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Answer 2

a)\(x^2-x\left(x+2\right)-1=2\\ \Rightarrow x^2-x^2-2x-1=2\\ \Rightarrow-2x=3\\ \Rightarrow x=-\dfrac{3}{2}\)

b) \(x^2=\left(2x-1\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=2x-1\\x=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Answer 3

a. x² - x(x + 2) - 1 = 2

<=> x² - x² - 2x = 3

<=> -2x = 3

<=> \(x=-\dfrac{3}{2}\)

b. x² = (2x - 1)²

<=> x² - (2x - 1)² = 0

<=> (x - 2x + 1)(x + 2x - 1) = 0

<=> (1 - x)(3x - 1) = 0

<=> \(\left[{}\begin{matrix}1-x=0\\3x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)