f: Ta có: \(x^2-10x+25=\left(2x-1\right)^2\)
\(\Leftrightarrow\left(2x-1-x+5\right)\left(2x-1+x-5\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
a) ĐKXĐ: \(x\ge0\)
\(\sqrt{x^2+x}=x\)
\(\Rightarrow x^2+x=x^2\Leftrightarrow x=0\) (tm)
Vậy pt có tập nghiệm \(S=\left\{0\right\}\)
b)ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le1\\x\ge1\end{matrix}\right.\)
\(\Leftrightarrow x=1\) (tm)
Vậy pt có tập nghiệm \(S=\left\{1\right\}\)
e: Ta có: \(\sqrt{x^4-2x^2+1}=x-1\)
\(\Leftrightarrow\left|x^2-1\right|=x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=x-1\left(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\right)\\x^2-1=1-x\left(-1< x< 1\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-1-1+x=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=1\left(loại\right)\end{matrix}\right.\)
