b/ ĐKXĐ: ...

\(2x^3-2y^3+5x-5y=0\)

\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2\right)+5\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+5\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)^2+x^2+y^2+5\right]=0\)

\(\Leftrightarrow x=y\) (ngoặc sau luôn dương)

Thế vào pt dưới:

\(\frac{3x}{x^2+x+1}+\frac{5x}{x^2+3x+1}=2\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{3}{x+\frac{1}{x}+1}+\frac{5}{x+\frac{1}{x}+3}=2\)

Đặt \(x+\frac{1}{x}+1=t\)

\(\Rightarrow\frac{3}{t}+\frac{5}{t+2}=2\Leftrightarrow3\left(t+2\right)+5t=2t\left(t+2\right)\)

\(\Leftrightarrow2t^2-4t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)

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

a/ ĐKXĐ: ...

\(2x-\frac{1}{y}=2y-\frac{1}{x}\Leftrightarrow\frac{2xy-1}{y}=\frac{2xy-1}{x}\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\2xy-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\xy=\frac{1}{2}\end{matrix}\right.\)

TH1: \(x=y\Rightarrow6x^2=7x^2-8\Rightarrow x^2=8\Rightarrow...\)

TH2: \(xy=\frac{1}{2}\Rightarrow y=\frac{1}{2x}\)

\(\Rightarrow2\left(2x^2+\frac{1}{4x^2}\right)+4\left(x-\frac{1}{2x}\right)=\frac{7}{2}-8\)

\(\Leftrightarrow4\left(x^2+\frac{1}{4x^2}\right)+8\left(x-\frac{1}{2x}\right)+9+4x^2=0\)

Đặt \(x-\frac{1}{2x}=t\Rightarrow x^2+\frac{1}{4x^2}=t^2+1\)

\(\Rightarrow4\left(t^2+1\right)+8t+9+4x^2=0\)

\(\Leftrightarrow4\left(t+1\right)^2+4x^2+9=0\)

Vế trái luôn dương nên pt vô nghiệm

Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{x+y}{xy}=2\\ (x+y)^2-2xy=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+y=2xy\\ (x+y)^2-2xy=2\end{matrix}\right.\)

\(\Rightarrow (2xy)^2-2xy=2\)

\(\Leftrightarrow 2(xy)^2-xy-1=0\)

\(\Leftrightarrow 2xy(xy-1)+(xy-1)=0\Leftrightarrow (xy-1)(2xy+1)=0\)

\(\Leftrightarrow \left[\begin{matrix} xy=1\\ xy=\frac{-1}{2}\end{matrix}\right.\)

Nếu $xy=1\Rightarrow x+y=2xy=2$

$\Rightarrow y=2-x\Rightarrow xy=x(2-x)=1$

$\Leftrightarrow x^2-2x+1=0\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\Rightarrow y=\frac{1}{x}=1$

Nếu $xy=\frac{-1}{2}\Rightarrow x+y=2xy=-1$

$\Rightarrow y=-1-x\Rightarrow xy=x(-1-x)=\frac{-1}{2}$

$\Leftrightarrow x^2+x-\frac{1}{2}=0\Rightarrow x=\frac{-1+\sqrt{3}}{2}$

$\Rightarrow y=\frac{-1}{2x}=\frac{-1\mp \sqrt{3}}{2}$

Vậy $(x,y)=(1,1); (\frac{-1+\sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}); (\frac{-1-\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2})$

Ta có:

Đặt \(\left\{{}\begin{matrix}\frac{1}{x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\)

hpt \(\Leftrightarrow\left\{{}\begin{matrix}2a+6b=1,1\\4a-9b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{9b+1}{4}\\\frac{2\cdot\left(9b+1\right)}{4}-9b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{-1}{9}\\a=\frac{9b+1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=\frac{-1}{9}\end{matrix}\right.\)

Pt vô nghiệm.