Question

X²+(14-x)²=100

Answer 1

Ta có: \(x^2+\left(14-x\right)^2=100\)

\(\Leftrightarrow x^2+x^2-28x+196-100=0\)

\(\Leftrightarrow2x^2-28x+96=0\)

\(\Leftrightarrow2x^2-16x-12x+96=0\)

\(\Leftrightarrow2x\left(x-8\right)-12\left(x-8\right)=0\)

\(\Leftrightarrow2\left(x-8\right)\left(x-6\right)=0\)

mà 2≠0

nên \(\left[{}\begin{matrix}x-8=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=6\end{matrix}\right.\)

Vậy: S={8;6}

Answer 2

\(x^2+\left(14-x\right)^2=100\\ \Leftrightarrow x^2+x^2-28x+196-100=0\\ \Leftrightarrow2x^2-28x+96=0\\ \Leftrightarrow2x^2-12x-16x+96=0\\ \Leftrightarrow2x\left(x-6\right)-16\left(x-6\right)=0\\ \Leftrightarrow\left(2x-16\right)\left(x-6\right)=0\\ \Leftrightarrow2\left(x-8\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=6\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{8;6\right\}\)