Question

Tìm x

\(\sqrt{x+1}=x-5\)

Thắng, bài này nữa :D

Answer 1

ôi cha dễ quá BÌnh thẳng lên thôi

\(\sqrt{\left(x+1\right)^2}=\left(x-5\right)^2\)

\(\Leftrightarrow x+1=x^2-10x+25\)

\(\Leftrightarrow-x^2+11x-24=0\)

\(\Leftrightarrow-\left(x^2-11x+24\right)=0\)

\(\Leftrightarrow x^2-8x-3x+24=0\)

\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=8\end{cases}\left(tm\right)}\)