Giải phương trình
1) \(\sqrt{x^2+10x+21}\)+6=3\(\sqrt{x+3}\)+2\(\sqrt{x+7}\)
2) \(\sqrt{x}\)+\(\sqrt{x-5}\)+\(\sqrt{x+7}\)=9
3) (x-2)(x+1)+3(x-2)\(\sqrt{\frac{x+1}{x-2}}\)=10
4) 3x+7\(\sqrt{x-4}\)=14\(\sqrt{x-4}\)-20
5)\(\left(\sqrt{x+1}+1\right)\)\(\left(\sqrt{x+1}+2x-5\right)\)=x
6)\(\sqrt{x^2-3x+2}\)+\(\sqrt{x+3}\)=\(\sqrt{x-2}\)+\(\sqrt{x^2+2x-3}\)
1.
ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+7\right)}-3\sqrt{x+3}-2\left(\sqrt{x+7}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+7}-3\right)-2\left(\sqrt{x+7}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+3}-2\right)\left(\sqrt{x+7}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{x+7}=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2.
\(\Leftrightarrow\sqrt{x}-3+\sqrt{x-5}-2+\sqrt{x+7}-4=0\)
\(\Leftrightarrow\frac{x-9}{\sqrt{x}+3}+\frac{x-9}{\sqrt{x-5}+2}+\frac{x-9}{\sqrt{x+7}+4}=0\)
\(\Leftrightarrow\left(x-9\right)\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x-5}+2}+\frac{1}{\sqrt{x+7}+4}\right)=0\)
\(\Leftrightarrow x=9\)
3.
ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)
- Với \(x>2\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)+3\sqrt{\frac{\left(x-2\right)^2\left(x+1\right)}{x-2}}-10=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)+3\sqrt{\left(x-2\right)\left(x+1\right)}-10=0\)
Đặt \(\sqrt{\left(x-2\right)\left(x+1\right)}=t\ge0\)
\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x-2\right)\left(x+1\right)}=2\Leftrightarrow\left(x-2\right)\left(x+1\right)=4\Leftrightarrow...\)
- Với \(x\le-1\Rightarrow x-2< 0\Rightarrow x-2=-\sqrt{\left(x-2\right)^2}\)
Pt tương đương:
\(\left(x-2\right)\left(x+1\right)-3\sqrt{\frac{\left(x-2\right)^2\left(x+1\right)}{x-2}}-10=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)-3\sqrt{\left(x-2\right)\left(x+1\right)}-10=0\)
Đặt \(\sqrt{\left(x-2\right)\left(x+1\right)}=t\ge0\Rightarrow t^2-3t-10=0\Rightarrow t=5\)
\(\Rightarrow\left(x-2\right)\left(x+1\right)=25\Leftrightarrow...\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
