Bài 2 : Giải các phương trình sau
1 , \(x\left(x+5\right)=2\sqrt[3]{x^2+5x-2}-2\)
2 , \(\sqrt[3]{x+5}+\sqrt[3]{x+6}=\sqrt[3]{2x+11}\)
3 , \(\sqrt[4]{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
4 , \(x^2-2x-8=4\sqrt{\left(4-x\right)\left(x+2\right)}\)
5 , \(x^2+5x+2+2\sqrt{x^2+5x+10}=0\)
6 , \(\sqrt{2x^2+3x-5}=x+1\)
7 , \(\left(x-1\right)\left(x-3\right)+3\sqrt{x^2-4x+5}-2=0\)
1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)
\(t^3+2=2t-2\)
\(\Leftrightarrow t^3-2t+4=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)
\(\Rightarrow t=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)
3/ ĐKXĐ: \(\left|x\right|\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{x-\sqrt{x^2-1}}=a>0\\\sqrt[4]{x+\sqrt{x^2-1}}=b>0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}ab=1\\a+b^2=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}ab=1\\a=2-b^2\end{matrix}\right.\)
\(\Rightarrow b\left(2-b^2\right)=1\Leftrightarrow b^3-2b+1=0\)
\(\Leftrightarrow\left(b-1\right)\left(b^2+b-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=1\\b^2+b-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}b=1\\b=\frac{-1+\sqrt{5}}{2}\\b=\frac{-1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[4]{x+\sqrt{x^2-1}}=1\\\sqrt[4]{x+\sqrt{x^2-1}}=\frac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x^2-1}=1\\x+\sqrt{x^2-1}=\frac{7-3\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2-1}=\frac{7-3\sqrt{5}}{2}-x\left(vn\right)\end{matrix}\right.\)
4/ ĐKXĐ: \(-2\le x\le4\)
Đặt \(\sqrt{\left(4-x\right)\left(x+2\right)}=t\ge0\Rightarrow x^2-2x-8=-t^2\)
\(\Leftrightarrow-t^2=4t\)
\(\Leftrightarrow t^2+4t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(4-x\right)\left(x+2\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
5/ Đặt \(\sqrt{x^2+5x+10}=t\ge0\)
\(t^2-8+2t=0\)
\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+10}=2\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
6/ \(x\ge-1\)
\(\Leftrightarrow2x^2+3x-5=\left(x+1\right)^2\)
\(\Leftrightarrow2x^2+3x-5=x^2+2x+1\)
\(\Leftrightarrow x^2+x-6=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\left(l\right)\end{matrix}\right.\)
7/ \(\Leftrightarrow x^2-4x+5+3\sqrt{x^2-4x+5}-4=0\)
Đặt \(\sqrt{x^2-4x+5}=t>0\)
\(\Rightarrow t^2+3t-4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-4x+5}=1\)
\(\Leftrightarrow x^2-4x+4=0\Rightarrow x=2\)