`sqrt{3-x}+sqrt{x+2}=3(1)`
`ĐKXĐ:-2<=x<=3`
`(1)<=>sqrt{3-x}-2+sqrt{x+2}-1=0`
`<=>(-x-1)/(sqrt{3-x}+2)+(x+1)/(sqrt{x+2}+1)=0`
`<=>(x+1)(1/(sqrt{x+2}+1)-1/(sqrt{3-x}+2))=0`
`=>` \(\left[ \begin{array}{l}x=-1(TM)\\x=\dfrac{1}{\sqrt{x+2}+1}=\dfrac{1}{\sqrt{3-x}+2}(2)\end{array} \right.\)
`giải` `(2)`
`<=>sqrt{x+2}+1=sqrt{3-x}+2`
`<=>sqrt{x+2}=sqrt{3-x}+1`
`text{ĐK để phương trình có nghiệm}`
`sqrt{x+2}>=1`
`=>x+2>=1`
`=>x>=-1`
`<=>x+2=3-x+1+2sqrt{3-x}`
`<=>2x-2=2sqrt{3-x}`
`<=>x-1=sqrt{3-x}`
`text{ĐK để phương trình có nghiệm}`
`x>=1`
`text{kết hợp ĐKXĐ}` `1<=x<=3`
`<=>x^2-2x+1=3-x`
`<=>x^2-x-2=0`
`<=>x^2+x-2x-2=0`
`<=>x(x+1)-2(x+1)=0`
`<=>(x+1)(x-2)=0`
`=>` \(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-1(loại)\\x=2(TM)\end{array} \right.\)
