b: ĐKXĐ: \(\begin{cases}x^2-4\ge0\\ x+2\ge0\end{cases}\)

=>\(x^2-4\ge0\) và x>=-2

=>(x-2)(x+2)>=0 và x>=-2

=>(x>=2 hoặc x<=-2) và x>=-2

=>x=-2 hoặc x>=2

Ta có: \(\sqrt{x^2-4}+\sqrt{x+2}=0\)

=>\(\sqrt{x+2}\left(\sqrt{x-2}+1\right)=0\)

=>\(\sqrt{x+2}=0\)

=>x+2=0

=>x=-2(nhận)

c: ĐKXĐ: \(\begin{cases}x^2-5x+6\ge0\\ x+1\ge0\\ x-2\ge0\\ x^2-2x-3\ge0\end{cases}\)

=>(x-2)(x-3)>=0 và x>=2 và (x-3)(x+1)>=0

=>(x>=3 hoặc x<=2) và x>=2 và (x>=3 hoặc x<=-1)

=>x>=3 hoặc x=2 hoặc x<=-1

\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

=>\(\sqrt{\left(x-2\right)\left(x-3\right)}-\sqrt{\left(x-3\right)\left(x+1\right)}+\sqrt{x+1}-\sqrt{x-2}=0\)

=>\(\sqrt{x-3}\left(\sqrt{x-2}-\sqrt{x+1}\right)-\left(\sqrt{x-2}-\sqrt{x+1}\right)=0\)

=>\(\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)

=>\(\sqrt{x-3}-1=0\)

=>\(\sqrt{x-3}=1\)

=>x-3=1

=>x=4(nhận)

a) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)

     \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)

     \(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)

     \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)

  Vậy ...

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-3\right)}-\sqrt{x-2}=\sqrt{\left(x-3\right)\left(x+1\right)}-\sqrt{x+1}\)

=>\(\sqrt{x-2}\left(\sqrt{x-3}-1\right)-\sqrt{x+1}\left(\sqrt{x-3}-1\right)=0\)

=>\(\left(\sqrt{x-3}-1\right)\left(\sqrt{x-2}-\sqrt{x+1}\right)=0\)

=>x-3=1

=>x=4

1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

e: ĐKXĐ: 2<=x<=4

Ta có: \(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)

=>\(\sqrt{x-2}-1+\sqrt{4-x}-1=2x^2-5x-3\)

=>\(\frac{x-2-1}{\sqrt{x-2}+1}+\frac{4-x-1}{\sqrt{4-x}+1}=2x^2-6x+x-3\)

=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}\right)=\left(x-3\right)\left(2x+1\right)\)

=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}-2x-1\right)=0\)

=>x-3=0

=>x=3(nhận)

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

Câu a bạn coi lại đề

b. ĐKXĐ: \(x\ge0;x\ne1\)

\(\Leftrightarrow\dfrac{\sqrt{2x+1}+\sqrt{3x}}{1-x}=\dfrac{\sqrt{3x+2}}{1-x}\)

\(\Leftrightarrow\sqrt{2x+1}+\sqrt{3x}=\sqrt{3x+2}\)

\(\Leftrightarrow5x+1+2\sqrt{3x\left(2x+1\right)}=3x+2\)

\(\Leftrightarrow2\sqrt{6x^2+3x}=1-2x\) (\(x\le\dfrac{1}{2}\) )

\(\Leftrightarrow4\left(6x^2+3x\right)=4x^2-4x+1\)

\(\Leftrightarrow20x^2+16x-1=0\)

\(\Rightarrow x=\dfrac{-4+\sqrt{21}}{10}\)

Bạn xem lại đề câu a.