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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 \(x=0\)
a/ \(\text{ĐK: }....\Leftrightarrow x\le-3\text{ hoặc }x\ge0\)
+TH1: \(x\ge0\)
\(pt\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow x=0\text{ hoặc }\sqrt{x+1}+\sqrt{x+2}=\sqrt{x+3}\text{ (1)}\)
\(\left(1\right)\Leftrightarrow x+1+x+2+2\sqrt{\left(x+1\right)\left(x+2\right)}=x+3\)
\(\Leftrightarrow x+2\sqrt{\left(x+1\right)\left(x+2\right)}=0\text{ (vô nghiệm do }x\ge0\text{ nên }x+\sqrt{\left(x+1\right)\left(x+2\right)}>0\text{)}\)
\(+TH2:\text{ }x\le-3\)
\(pt\Leftrightarrow\sqrt{-x}\left(\sqrt{-x-1}+\sqrt{-x-2}-\sqrt{-x-3}\right)=0\)
\(\Leftrightarrow\sqrt{-x-1}+\sqrt{-x-2}=\sqrt{-x-3}\text{ }\left(do\text{ }x\le-3\Rightarrow\sqrt{-x}>\sqrt{3}\right)\)
\(\Leftrightarrow-x-1-x-2+2\sqrt{\left(-x-1\right)\left(-x-2\right)}=-x-3\)
\(\Leftrightarrow2\sqrt{\left(-x-1\right)\left(-x-2\right)}-x=0\text{ (vô nghiệm do }-x\ge3\text{)}\)
Vậy \(x=0\)
b/
\(\text{ĐK: }x\ge1\)
\(\text{Đặt }\sqrt{x-1}=t;\text{ }t\ge0\)
\(pt\text{ thành: }\left(t+1\right)^3+2t+t^2-1=0\)
\(\Leftrightarrow t^3+4t^2+5t=0\Leftrightarrow t\left(t^2+4t+5\right)=0\)
\(\Leftrightarrow t=0\vee t^2+4t+5=0\text{ (Vô nghiệm)}\)
\(pt\text{ đã cho }\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x=1\)