Question

Tìm x :

a, \(\left(2x+\dfrac{1}{2}\right)^2-\left(1-2x\right)^2=0\)

b, \(x.\left(x-2\right)+2-x=0\)

Answer 1

a, \(\left(2x+\dfrac{1}{2}\right)^2-\left(1-2x\right)^2=0\)

\(\Leftrightarrow\left[\left(2x+\dfrac{1}{2}\right)-\left(1-2x\right)\right]\cdot\left[\left(2x+\dfrac{1}{2}\right)+\left(1-2x\right)\right]=0\)

\(\Leftrightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\cdot\left(2x+\dfrac{1}{2}+1-2x\right)=0\)

\(\Leftrightarrow\left[\left(4x+-\dfrac{1}{2}\right)\right]-\dfrac{3}{2}=0\)

\(\Leftrightarrow4x+\left(-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow4x=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{8}\)

b, \(x.\left(x-2\right)+2-x=0\)

\(\Leftrightarrow x.\left(x-2\right)+\left(2-x\right)=0\)

\(\Leftrightarrow x.\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

Vậy x = 2 hoặc 1

Answer 2

\(\left(2x+\dfrac{1}{2}\right)^2-\left(1-2x\right)^2=0\\ \left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=0\\ \left(4x-\dfrac{1}{2}\right).\dfrac{3}{2}=0\\ \Rightarrow4x-\dfrac{1}{2}=0\\ 4x=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{8}\)

Answer 3

\(x\left(x-2\right)+2-x=0\\ x\left(x-2\right)-\left(x-2\right)=0\\ \left(x-2\right)\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)