12: ĐKXĐ: x>=4
ta có: \(\sqrt{x+3}-\sqrt{x-4}=4\)
=>\(\left(\sqrt{x+3}-\sqrt{x-4}\right)^2=4^2=16\)
=>\(x+3+x-4-2\cdot\sqrt{\left(x+3\right)\left(x-4\right)}=16\)
=>\(2x-1-2\sqrt{\left(x+3\right)\left(x-4\right)}=16\)
=>\(\sqrt{4\left(x+3\right)\left(x-4\right)}=2x-1-16=2x-17\)
=>\(\begin{cases}4\left(x+3\right)\left(x-4\right)=\left(2x-17\right)^2\\ 2x-17\ge0\end{cases}\)
=>\(\begin{cases}4\left(x^2-4x+3x-12\right)=4x^2-68x+289\\ x\ge\frac{17}{2}\end{cases}\Rightarrow\begin{cases}4\left(x^2-x-12\right)=4x^2-68x+289\\ x\ge\frac{17}{2}\end{cases}\)
=>\(\begin{cases}4x^2-4x-48=4x^2-68x+289\\ x\ge\frac{17}{2}\end{cases}\Rightarrow\begin{cases}-4x+68x=289+48\\ x\ge\frac{17}{2}\end{cases}\)
=>\(\begin{cases}64x=337\\ x\ge\frac{17}{2}\end{cases}\Rightarrow\) x∈∅
9: ĐKXĐ: x>=2
Ta có: \(\sqrt{3x-5}=3-\sqrt{x-2}\)
=>\(\sqrt{3x-5}-2=1-\sqrt{x-2}\)
=>\(\frac{3x-5-4}{\sqrt{3x-5}+2}=\frac{1-\left(x-2\right)}{1+\sqrt{x-2}}\)
=>\(\frac{3x-9}{\sqrt{3x-5}+4}=\frac{-\left(x-3\right)}{\sqrt{x-2}+1}\)
=>\(\left(x-3\right)\left(\frac{3}{\sqrt{3x-5}+4}+\frac{1}{\sqrt{x-2}+1}\right)=0\)
=>x-3=0
=>x=3(nhận)
6: ĐKXĐ: -5<=x<=5
Ta có: \(\sqrt{x+5}+\sqrt{5-x}=4\)
=>\(\sqrt{x+5}-3+\sqrt{5-x}-1=0\)
=>\(\frac{x+5-9}{\sqrt{x+5}+3}+\frac{5-x-1}{\sqrt{5-x}+1}=0\)
=>\(\left(x-4\right)\left(\frac{1}{\sqrt{x+5}+3}-\frac{1}{\sqrt{5-x}+1}\right)=0\)
=>x-4=0
=>x=4(nhận)
3: ĐKXĐ: x>=0
ta có: \(\sqrt{x+5}=1+\sqrt{x}\)
=>\(\sqrt{x+5}-\sqrt{x}=1\)
=>\(\left(\sqrt{x+5}-\sqrt{x}\right)^2=1^2=1\)
=>\(x+5+x-2\sqrt{x\left(x+5\right)}=1\)
=>\(2\sqrt{x\left(x+5\right)}=2x+5-1=2x+4=2\left(x+2\right)\)
=>\(\sqrt{x\left(x+5\right)}=x+2\)
=>\(\begin{cases}x+2\ge0\\ x\left(x+5\right)=\left(x+2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-2\\ x^2+5x=x^2+4x+4\end{cases}\)
=>x=4(nhận)
11: ĐKXĐ: x>=5
Ta có: \(\sqrt{x-5}-\sqrt{x+4}=2\)
mà \(\sqrt{x-5}<\sqrt{x+4}\left(x-5
nên phương trình vô nghiệm
8: ĐKXĐ: x>=-1
Ta có: \(\sqrt{x+1}+\sqrt{x+6}=5\)
=>\(\left(\sqrt{x+1}+\sqrt{x+6}\right)^2=5^2=25\)
=>\(x+1+x+6+2\sqrt{\left(x+1\right)\left(x+6\right)}=25\)
=>\(2\sqrt{\left(x+1\right)\left(x+6\right)}=25-2x-7=-2x+18\)
=>\(\sqrt{\left(x+1\right)\left(x+6\right)}=-x+9\)
=>\(\begin{cases}\left(x+1\right)\left(x+6\right)=\left(-x+9\right)^2\\ -x+9\ge0\end{cases}\Rightarrow\begin{cases}x^2+7x+6=x^2-18x+81\\ x\le9\end{cases}\)
=>\(\begin{cases}7x+18x=81-6\\ -1\le x\le9\end{cases}\Rightarrow\begin{cases}25x=75\\ -1\le x\le9\end{cases}\)
=>x=3(nhận)
5: ĐKXĐ: -3<=x<=10
Ta có: \(\sqrt{10-x}+\sqrt{x+3}=5\)
=>\(\sqrt{10-x}-2-3+\sqrt{x+3}=5-5=0\)
=>\(\frac{10-x-4}{\sqrt{10-x}+2}+\frac{x+3-9}{\sqrt{x+3}+3}=0\)
=>\(\frac{6-x}{\sqrt{10-x}+2}+\frac{x-6}{\sqrt{x+3}+3}=0\)
=>\(\left(x-6\right)\left(-\frac{1}{\sqrt{10-x}+2}+\frac{1}{\sqrt{x+3}+3}\right)=0\)
=>x-6=0
=>x=6(nhận)
2: ĐKXĐ: 0<=x<=2
Ta có: \(\sqrt{x}+\sqrt{2-x}=1\)
=>\(\left(\sqrt{x}+\sqrt{2-x}\right)^2=1^2=1\)
=>\(x+2-x+2\cdot\sqrt{x\left(2-x\right)}=1\)
=>\(2\sqrt{x\left(2-x\right)}=1-2=-1\) (vô lý)
=>Phương trình vô nghiệm
10: ĐKXĐ: x>=2
Ta có: \(\sqrt{x+1}-\sqrt{x-2}=1\)
=>\(\sqrt{x+1}-2+1-\sqrt{x-2}=1-2+1=0\)
=>\(\frac{x+1-4}{\sqrt{x+1}+2}+\frac{1-\left(x-2\right)}{1+\sqrt{x-2}}=0\)
=>\(\frac{x-3}{\sqrt{x+1}+2}-\frac{x-3}{\sqrt{x-2}+1}=0\)
=>(x-3)\(\left(\frac{1}{\sqrt{x+1}+2}-\frac{1}{\sqrt{x-2}+1}\right)\) =0
=>x-3=0
=>x=3(nhận)
7: ĐKXĐ: \(\begin{cases}x-3\ge0\\ 2-x\ge0\end{cases}\Rightarrow\begin{cases}x\ge3\\ x\le2\end{cases}\Rightarrow x\in\) ∅
=>Phương trình vô nghiệm
4: ĐKXĐ: \(\begin{cases}x-1\ge0\\ 3-x\ge0\end{cases}\Rightarrow\begin{cases}x\ge1\\ x\le3\end{cases}\Rightarrow1\le x\le3\)
Ta có: \(\sqrt{x-1}+\sqrt{3-x}=2\)
=>\(\sqrt{x-1}-1+\sqrt{3-x}-1=2-2=0\)
=>\(\frac{x-1-1}{\sqrt{x-1}+1}+\frac{3-x-1}{\sqrt{3-x}+1}=0\)
=>\(\frac{x-2}{\sqrt{x-1}+1}-\frac{x-2}{\sqrt{3-x}+1}=0\)
=>x-2=0
=>x=2(nhận)
1: ĐKXĐ: 0<=x<=1
Ta có: \(\sqrt{x}+\sqrt{1-x}=1\)
=>\(\left(\sqrt{x}+\sqrt{1-x}\right)^2=1\)
=>\(x+1-x+2\sqrt{x\cdot\left(1-x\right)}=1\)
=>\(1+2\sqrt{x\left(1-x\right)}=1\)
=>\(2\sqrt{x\left(1-x\right)}=0\)
=>\(\sqrt{x\left(1-x\right)}=0\)
=>x(1-x)=0
=>\(\left[\begin{array}{l}x=0\left(nhận\right)\\ x=1\left(nhận\right)\end{array}\right.\)
