1:

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

=>x-3=0 hoặc \(\sqrt{x+3}=2\)

=>x=3 hoặc x+3=4

=>x=1(loại) hoặc x=3(nhận)

2:

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

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

=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)

=>4(12x^2-16x+3x-4)=(7x-6)^2

=>49x^2-84x+36=48x^2-52x-16

=>-84x+36=-52x-16

=>-32x=-52

=>x=13/8

3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)

=>|x-5|=5-x

=>x-5<=0

=>x<=5

4: \(\Leftrightarrow\left|x-4\right|=x+2\)

=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)

=>x>=-2 và -8x+16=4x+4

=>x=1

Dùng liên hợp.

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

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

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

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

\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)

\(=3x-1\)

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

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

<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)

<=> \(x^2-3x+2=0\) phương trình bậc 2.

Em làm tiếp nhé!

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

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x+2}+\sqrt{x-1}=2\sqrt{x+3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x+1+2\sqrt{\left(x-1\right)\left(x+2\right)}=4\left(x+3\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\2\sqrt{\left(x-1\right)\left(x+2\right)}=2x+11\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\4\left(x-1\right)\left(x+2\right)=4x^2+44x+121\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\-40x=129\end{cases}}\Rightarrow x=-3\) (thỏa)

b)\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)

Đk:\(x\ge-\frac{1}{3}\)

\(pt\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1-\left(\frac{3}{5}x+1\right)=\sqrt{3x+1}-\left(\frac{3}{5}x+1\right)\)

\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}-\frac{3}{5}x=\frac{3x+1-\left(\frac{3}{5}x+1\right)^2}{\sqrt{3x+1}+\frac{3}{5}x+1}\)

\(\Leftrightarrow\frac{3x\left(5-\sqrt{3x+10}\right)}{5\sqrt{3x+10}}=\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}\)

\(\Leftrightarrow\frac{3x\cdot\frac{25-3x-10}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)

\(\Leftrightarrow\frac{3x\cdot\frac{-3\left(x-5\right)}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)

\(\Leftrightarrow x\left(x-5\right)\left(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}\right)=0\)

Dễ thấy: \(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}< 0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

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

\(=3x+6\sqrt{x}-\left(x-1\right)\)

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

\(=2x+6\sqrt{x}+1\)

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

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

\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)

\(=-x+8\sqrt{x}+1\)

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

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

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

\(=4x-3\sqrt{x}-3\)

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

\(=x-9-\left(2x-3\sqrt{x}-2\right)\)

\(=x-9-2x+3\sqrt{x}+2\)

\(=-x+3\sqrt{x}-7\)

\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)

\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)

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

\(=x-4-4x-6\sqrt{x}+4\)

\(=-3-6\sqrt{x}\)