Question

[Nguyễn Châu Mỹ Linh]

Giải phương trình:

a) \(\sqrt{5-x}+\sqrt{x-3}=\sqrt{2}\)

b) \(\sqrt{x^2-4}=2\sqrt{x-2}\)

c) \(\sqrt{5x+7}=\sqrt{2x+3}+\sqrt{3x+4}\)

d) \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a) ĐKXĐ: \(\left\{{}\begin{matrix}5-x\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x\ge-5\\x\ge3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ge3\end{matrix}\right.\Leftrightarrow3\le x\le5\)

Ta có: \(\sqrt{5-x}+\sqrt{x-3}=\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x-3}\right)^2=\left(\sqrt{2}\right)^2\)

\(\Leftrightarrow5-x+2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}+x-3=2\)

\(\Leftrightarrow2+2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}=2\)

\(\Leftrightarrow2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}=0\)

mà \(2\ne0\)

nên \(\sqrt{\left(5-x\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\left(5-x\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5-x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

Vậy: S={3;5}

b) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow x-2\ge0\)\(\Leftrightarrow x\ge2\)

Ta có: \(\sqrt{x^2-4}=2\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{x-2}\cdot\sqrt{x+2}-2\cdot\sqrt{x-2}=0\)

\(\Leftrightarrow\sqrt{x-2}\cdot\left(\sqrt{x+2}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\)

Vậy: S={2}