a/ ĐKXĐ: \(x^2+2x-6\ge0\)
\(\Leftrightarrow x^2+2x-6+\left(x-2\right)\sqrt{x^2+2x-6}=0\)
\(\Leftrightarrow\sqrt{x^2+2x-6}\left(\sqrt{x^2+2x-6}+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-6}=0\left(1\right)\\\sqrt{x^2+2x-6}=2-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+2x-6=0\Rightarrow x=-1\pm\sqrt{7}\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\x^2+2x-6=\left(2-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\6x=10\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{3}\)
Câu b nhìn ko ra hướng, ko biết đề có nhầm đâu ko :(
c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(x^2+x\right)\left(x^2+x+2\right)}-\left(3-x\right)\sqrt{x^2+x}=0\)
\(\Leftrightarrow\sqrt{x^2+x}\left(\sqrt{x^2+x+2}-3+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\left(1\right)\\\sqrt{x^2+x+2}=3-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)
d/
Ta có \(\sqrt{x^2+3x+4}=\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{7}{4}}>1\)
\(\Rightarrow\sqrt{x^2+3x+4}-1>0\)
Nhân 2 vế của pt với \(\sqrt{x^2+3x+4}-1\)
\(\left(\sqrt{x^2+3x+4}-1\right)\left(x^2+3x+3\right)=3x\left(x^2+3x+3\right)\)
\(\Leftrightarrow\left(x^2+3x+3\right)\left(\sqrt{x^2+3x+4}-1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+3=0\left(vn\right)\\\sqrt{x^2+3x+4}=3x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)
e/ ĐKXĐ: \(3x^2-9x+1\ge0\)
\(\Leftrightarrow3x^2-9x+1-x^2=2\left(\sqrt{3x^2-9x+1}+x\right)\)
\(\Leftrightarrow\left(\sqrt{3x^2-9x+1}+x\right)\left(\sqrt{3x^2-9x+1}+x\right)=2\left(\sqrt{3x^2-9x+1}+x\right)\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{3x^2-9x+1}+x=0\left(1\right)\\\sqrt{3x^2-9x+1}-x=2\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{3x^2-9x+1}=-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\3x^2-9x+1=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le0\\2x^2-9x+1=0\end{matrix}\right.\) \(\Rightarrow x=\frac{9\pm\sqrt{73}}{4}\left(l\right)\)
\(\left(2\right)\Leftrightarrow\sqrt{3x^2-9x+1}=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\3x^2-9x+1=\left(x+2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\2x^2-13x-3=0\end{matrix}\right.\)
\(\Rightarrow x=\frac{13\pm\sqrt{193}}{4}\)
b/ ĐKXĐ: \(x^2+2x-2\ge0\)
\(\Leftrightarrow x^3-8-7x\sqrt{x^2+2x-2}+14\sqrt{x^2+2x-2}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-7\sqrt{x^2+2x-2}\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-7\sqrt{x^2+2x-2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x^2+2x+4-7\sqrt{x^2+2x-2}=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(\sqrt{x^2+2x-2}=a\ge0\Rightarrow x^2+2x=a^2+2\)
\(\left(1\right)\Leftrightarrow a^2+2+4-7a=0\)
\(\Leftrightarrow a^2-7a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-2}=1\\\sqrt{x^2+2x-2}=6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2+2x-2=1\\x^2+2x-2=36\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x-3=0\\x^2+2x-38=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=-1\pm\sqrt{39}\end{matrix}\right.\)