Question

Answer 1

\(a,A=\left(x+1\right)^2+\left(x-1\right)^2-\left(x-1\right)\left(x+1\right)\\ =x^2+2x+1+x^2-2x+1-x^2+1\\ =x^2+3\)

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

Vậy   \(x=0;x=1\)

Answer 2

a)

\(\left(x+1\right)^2+\left(x-1\right)^2-\left(x-1\right)\left(x+1\right)\)

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

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

\(=x^2+3\)

b)

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

Answer 3

\(a,A=\left(x+1\right)^2+\left(x-1\right)^2-\left(x-1\right)\left(x+1\right)\)
\(A=x^2+2x+1+x^2-2x+1-x^2+1\)
\(A=x^2+3\)
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\(b,x^2-x=0\)
\(\Rightarrow x\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)            \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{0;1\right\}\)