Question

[Lan Ngọc Ninh Dương]

Giải các phương trình

a,\(\sqrt{x}+\sqrt{x+9}=3\)

b,\(\sqrt{x\left(x-1\right)}=\sqrt{x\left(x+2\right)}\)

c,\(\sqrt{x-1}+\sqrt{x-7}=4\)

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

Answer 1

[Hồng Phúc]

a, ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+9+2\sqrt{x^2+9x}=9\)

\(\Leftrightarrow\sqrt{x^2+9x}=-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\\x^2+9x=x^2\end{matrix}\right.\Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ: \(x=0;x\le-2;x\ge1\)

\(pt\Leftrightarrow x\left(x-1\right)=x\left(x+2\right)\)

\(\Leftrightarrow-3x=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

c, ĐKXĐ: \(x\ge7\)

\(pt\Leftrightarrow2x-8+2\sqrt{\left(x-1\right)\left(x-7\right)}=16\)

\(\Leftrightarrow\sqrt{x^2-8x+7}=12-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}12-x\ge0\\x^2-8x+7=\left(12-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le12\\16x=137\end{matrix}\right.\Leftrightarrow x=\frac{137}{16}\left(tm\right)\)

d, ĐKXĐ: \(0\le x\le1\)

\(pt\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\\sqrt{x+2}+\sqrt{x-1}=\sqrt{x-2}\end{matrix}\right.\)

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

\(\Leftrightarrow2x+1+2\sqrt{\left(x+2\right)\left(x-1\right)}=x-2\)

\(\Leftrightarrow2\sqrt{x^2+x-2}=-x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x-3\ge0\\4\left(x^2+x-2\right)=\left(-x-3\right)^2\end{matrix}\right.\)

\(\Rightarrow x\le-3\left(\text{trái với ĐKXĐ}\right)\)

Vậy phương trình đã cho có nghiệm \(x=0\)