a) \(Q^2=2x-\sqrt{\left(x^2+2x+1\right)}\)
\(Q^2=2x-\sqrt{\left(x+1\right)^2}\)
\(Q^2=2x-x-1\)
\(Q^2=x-1\)
\(Q=\sqrt{x-1}\)
b) Để Q=9 thì \(\sqrt{x-1}=9\) => \(x=82\)
Mình lộn câu b) Để Q=7 =>\(\sqrt{x-1}\)=7 => x=50
a: Ta có: \(Q=2x-\sqrt{x^2+2x+1}\)
\(=2x-\left|x+1\right|\)
\(=\left[{}\begin{matrix}2x-x-1=x-1\left(x\ge-1\right)\\2x+x+1=3x+1\left(x< -1\right)\end{matrix}\right.\)
b: Ta có: Q=7
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\left(x\ge-1\right)\\3x+1=7\left(x< -1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(loại\right)\\x=2\left(loại\right)\end{matrix}\right.\)
