a) dat x-1=a

x=a+1

\(a+1+\sqrt{5+\sqrt{a}}=6\)

\(5-a=\sqrt{5+\sqrt{a}}\)

\(25-10a+a^2=5+\sqrt{a}\)

\(20-10a+a^2-\sqrt{a}=0\)

(a - \sqrt{5} - 5) (a + \sqrt{a} - 4) = 0

đúng nhưng b,c,d đâu

ý c)  dk tu viet

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

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

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

\(2x+2=4\)

2x=2

x=1

\(\sqrt{x^2-x+1}+\sqrt{x^2-9x+9}=2x\)

=>\(\sqrt{x^2-x+1}-x+\sqrt{x^2-9x+9}-x=0\)

=>\(\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}+\dfrac{x^2-9x+9-x^2}{\sqrt{x^2-9x+9}+x}=0\)

=>\(\left(-x+1\right)\left(\dfrac{1}{\sqrt{x^2-x+1}+x}+\dfrac{9}{\sqrt{x^2-9x+9}+x}\right)=0\)

=>-x+1=0

=>x=1

ĐK: \(x^3+4x^2+5x+6\ge0\)

Ta có: \(x^3+4x^2+5x+6=\left(x+3\right)\left(x^2+x+2\right);x^2+2x+5=\left(x+3\right)+\left(x^2+x+2\right)\)

Đặt \(\hept{\begin{cases}\sqrt{x+3}=u\\\sqrt{x^2+x+2}=v\end{cases}}\)

Vậy nên ta có phương trình: \(\)\(u^2+v^2=\frac{5}{2}uv\)

\(\Leftrightarrow2u^2-5uv+2v^2=0\Leftrightarrow\orbr{\begin{cases}u=2v\\u=\frac{1}{2}v\end{cases}}\)

Với u = 2v ta có: \(\sqrt{x+3}=2\sqrt{x^2+x+2}\Leftrightarrow x+3=4x^2+4x+8\)

\(\Leftrightarrow4x^2+3x+5=0\)   (Vô nghiệm)

Với \(u=\frac{1}{2}v\) ta có: \(2\sqrt{x+3}=\sqrt{x^2+x+2}\Leftrightarrow4x+12=x^2+x+2\)

\(\Leftrightarrow x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\left(tmđk\right)\)

Vậy phương trình có nghiệm \(x\in\left\{5;-2\right\}\)