Question

Giải pt

a) \(\sqrt{x}\) =2

b) \(\sqrt{25x-25}\) -10=0

c) \(\sqrt{25-10+x^2}\) =7

Answer 1

a)ĐKXĐ: \(x\ge0\)

Ta có: \(\sqrt{x}=2\)

\(\Leftrightarrow x=2^2=4\)(nhận)

Vậy: S={4}

b) ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{25x-25}-10=0\)

\(\Leftrightarrow\sqrt{25}\cdot\sqrt{x-1}=10\)

\(\Leftrightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=2^2=4\)

hay x=5(nhận)

Vậy: S={5}

c)ĐKXĐ: \(x\in Z\)

Ta có: \(\sqrt{25-10x+x^2}=7\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7\)

\(\Leftrightarrow\left|x-5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={12;-2}