a. ĐKXĐ \(x\ge2\)

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

\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x^2-x-1=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^2-x-1=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x=2\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\) Pt vô nghiệm 

\(a.\sqrt{x+3}=5-\sqrt{x-2}\)

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

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

\(x+3+x-2=25\)

\(2x+1=25\)

\(x=12\)

\(b.\sqrt{x^2-x-1}=1-x\)

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

\(x^2-x-1=1-2x+x^2\)

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

\(x-2=0\)

\(x=2\)

ĐKXĐ: x+3>=0

=>x>=-3

\(x+\left(x+1\right)\sqrt{x+3}=5\)

=>\(x+\sqrt{\left(x+3\right)\left(x+1\right)^2}=5\)

=>\(x+\sqrt{\left(x+3\right)\left(x^2+2x+1\right)}=5\)

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

=>\(x+\sqrt{x^3+5x^2+7x+3}=5\)

=>\(x-1+\sqrt{x^3+5x^2+7x+3}-4=0\)

=>\(\left(x-1\right)+\dfrac{x^3+5x^2+7x+3-16}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)+\dfrac{x^3-x^2+6x^2-6x+13x-13}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)+\dfrac{\left(x-1\right)\left(x^2+6x+13\right)}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)\left(1+\dfrac{x^2+6x+13}{\sqrt{x^3+5x^2+7x+3}+4}\right)=0\)

=>x-1=0

=>x=1(nhận)

\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)

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

\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

\(\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0\)

\(\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0\)

\(\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6+4\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{1;6+4\sqrt{5}\right\}\)