ĐKXĐ: \(x\ge1\)

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

Đặt \(\sqrt{x-\sqrt{x^2-1}}=t\Rightarrow\sqrt{x+\sqrt{x^2-1}}=\dfrac{1}{t}\)

Phương trình trở thành:

\(t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\Rightarrow t=1\)

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

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

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

\(\Rightarrow x=1\) (thỏa mãn)

Đặt \(\dfrac{x}{\sqrt{4x-1}}=a\)

Theo đề, ta có phương trình:

a+1/a=2

\(\Leftrightarrow a+\dfrac{1}{a}=2\)

\(\Leftrightarrow\dfrac{a^2+1-2a}{a}=0\)

=>a=1

=>\(x=\sqrt{4x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4x-1\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=3\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow x\in\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)

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

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

\(\Leftrightarrow x^2-7-\left(x\sqrt{x^2+2}-3x\right)=2\sqrt{x^2+2}-6\)

\(\Leftrightarrow x^2-7-\dfrac{x^2\left(x^2+2\right)-9x^2}{x\sqrt{x^2+2}+3x}=\dfrac{4\left(x^2+2\right)-36}{2\sqrt{x^2+2}+6}\)

\(\Leftrightarrow x^2-7-\dfrac{x^4-7x^2}{x\sqrt{x^2+2}+3x}-\dfrac{4x^2-28}{2\sqrt{x^2+2}+6}=0\)

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

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

Dễ thấy: \(1-\dfrac{x^2}{x\sqrt{x^2+2}+3x}-\dfrac{4}{2\sqrt{x^2+2}+6}>0\)

\(\Rightarrow x^2-7=0\Rightarrow x=\pm\sqrt{7}\)

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

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

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

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

=>-x+1=0

=>x=1

a)\(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)

\(\Leftrightarrow3\left(\dfrac{2x^2+1-1}{\sqrt{2x^2+1}+1}\right)-x\left(1+3x+8\sqrt{2x^2+1}\right)=0\)

\(\Leftrightarrow\dfrac{6x^2}{\sqrt{2x^2+1}+1}-x\left(1+3x+8\sqrt{2x^2+1}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{6x}{\sqrt{2x^2+1}+1}-\left(1+3x+8\sqrt{2x^2+1}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{6x}{\sqrt{2x^2+1}+1}=1+3x+8\sqrt{2x^2+1}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{2x^2+1}\\b=3x\end{matrix}\right.\left(a>0\right)\) thì

\(pt\left(2\right)\Leftrightarrow\)\(\dfrac{2b}{a+1}=1+b+8a\)

\(\Rightarrow\left\{{}\begin{matrix}a=-17\\b=120\end{matrix}\right.;\left\{{}\begin{matrix}a=-8\\b=49\end{matrix}\right.;\left\{{}\begin{matrix}a=-5\\b=26\end{matrix}\right.;\left\{{}\begin{matrix}a=-2\\b=5\end{matrix}\right.;\left\{{}\begin{matrix}a=-0\\b=1\end{matrix}\right.\) (loại vì \(a>0\))

Hay pt vô nghiệm

phần a liên hợp nhưng cx có yếu tố đặt ẩn là done r` nhé ;v còn phần b dg nghĩ có lẽ liên hợp nốt mà chủ thớt khó quá:v

e có 1 cách ngoài liên hợp cho câu b, rất đơn giản( nhưng dễ nhầm ) , đó là lập phương liên tiếp :v =))

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