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

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)\(\left|x-2y\right|=5\Rightarrow\left[\begin{matrix}x-2y=5\\x-2y=-5\end{matrix}\right.\)

Từ \(2x=3y=5z\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)\(\Rightarrow\frac{x}{15}=\frac{2y}{20}=\frac{z}{6}\)

Nếu x-2y=5

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{2y}{20}=\frac{z}{6}=\frac{x-2y}{15-20}=\frac{5}{-5}-1\)

\(\Rightarrow\left\{\begin{matrix}x=-15\\y=-10\\z=-6\end{matrix}\right.\)

Nếu x-2y=-5

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{2y}{20}=\frac{z}{6}=\frac{x-2y}{15-20}=\frac{-5}{-5}=1\)

\(\Rightarrow\left\{\begin{matrix}x=15\\y=10\\z=6\end{matrix}\right.\)

Vậy có 2 bộ (x,y,z). Đó là (-15;-10;-6), (15;10;6)

b) Từ \(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\)\(\Rightarrow\frac{x}{6}=\frac{y}{15}\left(1\right)\)

\(2x=3z\Rightarrow\frac{x}{3}=\frac{z}{2}\)\(\Rightarrow\frac{x}{6}=\frac{z}{4}\left(2\right)\)

Từ (1),(2)\(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{4}\)

Đặt\(\)\(\frac{x}{6}=\frac{y}{15}=\frac{x}{4}=k\)

\(\Rightarrow\left\{\begin{matrix}x=6k\\y=15k\\z=4k\end{matrix}\right.\Rightarrow xy=90k^2\)

\(\Rightarrow90k^2=90\Rightarrow k^2=1\Rightarrow\left[\begin{matrix}k=1\\k=-1\end{matrix}\right.\)

Với k=1\(\Rightarrow\)\(\left\{\begin{matrix}x=6\\y=15\\z=4\end{matrix}\right.\)

Với k=-1\(\Rightarrow\left\{\begin{matrix}x=-6\\y=-15\\z=-4\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)

\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)

\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)

\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)