ĐK: \(x\ge8\)

Đặt \(a=\sqrt[3]{x-1}\text{ (}a\ge\sqrt[3]{7}\text{)};\text{ }b=\sqrt{x-8}\text{ (}b\ge0\text{)}\Rightarrow x=b^2+8\)

\(a^3-b^2=x-1-\left(x-8\right)=7\text{ (*)}\)

\(pt\text{ thành }a^2-2a-\left(b^2+8-5\right)b-3\left(b^2+8\right)+31=0\)

\(\Leftrightarrow\left(a^2-2a\right)-\left(b^3+3b^2+3b\right)+7=0\)

\(\Leftrightarrow\left(a-1\right)^2-\left(b+1\right)^3+a^3-b^2=0\)

Đặt \(b+1=c\text{ (}c\ge1\text{)}\)

\(pt\text{ thành }a^3-c^3+\left(a-1\right)^2-\left(c-1\right)^2=0\)

\(\Leftrightarrow\left(a-c\right)\left(a^2+ac+c^2\right)+\left(a-c\right)\left(a+c-2\right)=0\)

\(\Leftrightarrow\left(a-c\right)\left[a^2+c^2+a+c+ac-2\right]=0\)

\(\Leftrightarrow a-c=0\text{ (do }a^2+c^2+a+c+ac-2>0\text{ với mọi }a\ge\sqrt[3]{7};c\ge1\text{)}\)

\(\Leftrightarrow a=c\Leftrightarrow a=b+1\)

Thay \(b=a-1\) vào \(\left(\text{*}\right)\)ta được

\(a^3-\left(a-1\right)^2=7\Leftrightarrow\left(a-2\right)\left(a^2+a+4\right)=0\)

\(\Leftrightarrow a-2=0\text{ hoặc }a^2+a+4=0\text{ (vô nghiệm)}\)

\(\Leftrightarrow a=2\)

\(\Rightarrow\sqrt[3]{x-1}=2\Leftrightarrow x=9\)

Kết luận: \(x=9\).

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

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

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

\(a,ĐK:x,y\ne2\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\y-2=b\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{3}{a}+\dfrac{2}{b}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{a}+\dfrac{9}{b}=15\\\dfrac{6}{a}+\dfrac{4}{b}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{5}{b}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+3=5\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow x=y=3\left(tm\right)\)

\(b,ĐK:x\ge3;y\ge1\)

Sửa: \(\sqrt{x-3}-\sqrt{y-1}=4\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\ge0\\b=\sqrt{y-1}\ge0\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a-2b=2\\a-b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\-b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3=36\\y-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=39\\y=5\end{matrix}\right.\)

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 =))

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

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

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

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

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

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

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

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

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