7: ĐKXĐ: \(\begin{cases}x-3\ge0\\ 2-x\ge0\end{cases}\Rightarrow\begin{cases}x\ge3\\ x\le2\end{cases}\Rightarrow2\le x\le3\)
Ta có: \(\sqrt{x-3}+\sqrt{2-x}=5\)
=>\(\left(\sqrt{x-3}+\sqrt{2-x}\right)^2=5^2=25\)
=>\(x-3+2-x+2\cdot\sqrt{\left(x-3\right)\left(2-x\right)}=25\)
=>\(2\sqrt{\left(x-3\right)\left(2-x\right)}-1=25\)
=>\(2\sqrt{\left(x-3\right)\left(2-x\right)}=26\)
=>\(\sqrt{\left(x-3\right)\left(2-x\right)}=13\)
=>(x-3)(2-x)=169
=>(x-3)(x-2)=-169
=>\(x^2-5x+6+169=0\)
=>\(x^2-5x+175=0\)
\(\Delta=\left(-5\right)^2-4\cdot1\cdot175=25-700=-675<0\)
=>Phương trình vô nghiệm
8: ĐKXĐ: x>=-1
Ta có: \(\sqrt{x+1}+\sqrt{x+6}=5\)
=>\(\sqrt{x+1}-2+\sqrt{x+6}-3=5-5=0\)
=>\(\frac{x+1-4}{\sqrt{x+1}+2}+\frac{x+6-9}{\sqrt{x+6}+3}=0\)
=>\(\frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
=>x-3=0
=>x=3(nhận)
10: ĐKXĐ: x>=2
Ta có: \(\sqrt{x+1}-\sqrt{x-2}=1\)
=>\(\sqrt{x+1}-2+1-\sqrt{x-2}=1-2+1=2-2=0\)
=>\(\frac{x+1-4}{\sqrt{x+1}+2}+\frac{1-\left(x-2\right)}{1+\sqrt{x-2}}=0\)
=>\(\frac{x-3}{\sqrt{x+1}+2}-\frac{x-3}{\sqrt{x-2}+1}=0\)
=>x-3=0
=>x=3(nhận)
11: ĐKXĐ: x>=5
Ta có: \(\sqrt{x-5}-\sqrt{x+4}=2\)
=>\(\left(\sqrt{x-5}-\sqrt{x+4}\right)^2=2^2=4\)
=>\(x-5+x+4-2\sqrt{\left(x-5\right)\left(x+4\right)}=4\)
=>\(2\sqrt{\left(x-5\right)\left(x+4\right)}=2x-1-4=2x-5\)
=>\(\sqrt{4\left(x-5\right)\left(x+4\right)}=2x-5\)
=>\(\begin{cases}2x-5\ge0\\ 4\left(x-5\right)\left(x+4\right)=2x-5\end{cases}\Rightarrow\begin{cases}x\ge\frac52\\ 4\left(x^2-x-20\right)=2x-5\end{cases}\)
=>\(\begin{cases}x\ge5\\ 4x^2-4x-80-2x+5=0\end{cases}\Rightarrow\begin{cases}x\ge5\\ 4x^2-6x-75=0\end{cases}\)
Ta có: \(4x^2-6x-75=0\) (1)
\(\Delta=\left(-6\right)^2-4\cdot4\cdot\left(-75\right)=1236>0\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{6-\sqrt{1236}}{2\cdot4}=\frac{6-2\sqrt{309}}{8}=\frac{3-\sqrt{309}}{4}\left(loại\right)\\ x=\frac{6+\sqrt{1236}}{2\cdot4}=\frac{6+2\sqrt{309}}{8}=\frac{3+\sqrt{309}}{4}\left(nhận\right)\end{array}\right.\)
