1, bình phương x rồi rút gọn ta được

\(x^2=3\sqrt{10}-4\sqrt{2}-2\sqrt{2}.\sqrt{2\left(\sqrt{5}-1\right)\left(\sqrt{5}-2\right)}\)

=\(3\sqrt{10}-4\sqrt{2}-2\sqrt{2}.\sqrt{14-6\sqrt{5}}\)

=\(3\sqrt{10}-4\sqrt{2}-2\sqrt{2}.\sqrt{\left(3-\sqrt{5}\right)^2}\)

=\(3\sqrt{10}-4\sqrt{2}-2\sqrt{2}\left(3-\sqrt{5}\right)\)

=\(5\sqrt{10}-10\sqrt{2}>0\)

=>x=\(\sqrt{5\sqrt{10}-10\sqrt{2}}\)

4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)

\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)

\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

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

\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)

Chắc tới đây bạn làm đc rồi nhỉ

b,c đề ko ổn

đm m lm lắm thế 

a)ĐK:\(-1\le x\le1\)

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

\(\Leftrightarrow1+\sqrt{1-x^2}=x^2+4x^2\left(1-x^2\right)+4x^2\sqrt{1-x^2}\)

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

SUy ra x=1/2 và pt trong ngoặc suy ra x=1 (bn làm tiếp) 

c)\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{10+3x-x^2}=4\)

\(\Leftrightarrow\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(5-x\right)\left(x+2\right)}=4\)

Đặt \(\sqrt{x+2}=a;\sqrt{5-x}=b\left(a,b\ge0\right)\):

\(a+b+ab=4\)\(\Leftrightarrow\left(a+1\right)\left(b+1\right)=3\)

Ok tiếp nhé