\(\hept{\begin{cases}\frac{2x-3y}{2y-5}=\frac{3x+1}{3y-4}\left(1\right)\\2\left(x-3\right)-3\left(y+2\right)=-16\left(2\right)\end{cases}}\)

Nhân chéo và chuyển vế phương trình (1) và nhân phân phối, chuyển vế phương trình (2), ta được:

\(\hept{\begin{cases}7x-11y=-17\\2x-3y=-4\end{cases}}\)

<=>\(\hept{\begin{cases}x=7\\y=6\end{cases}}\)

\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{x^2+2x-3}\)
\(ĐKXĐ:x^2+2x-3=\left(x+1\right)\left(x-3\right)\\ \Rightarrow x\ne-1;x\ne3\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x^2\)
\(\Leftrightarrow x^2+x+x^2-3x=4x^2\)
\(\Leftrightarrow2x^2-2x=4x^2\)
\(\Leftrightarrow2x^2-4x^2-2x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow2x\left(-x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\-x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(N\right)\\x=-1\left(L\right)\end{cases}}\)
Tự kết luận tập nghiệm bạn nhé!

 x²+2x-3 = (x+1)(x-3)

vậy MSC = 2(X+1(X-3) qui đồng mẫu số r làm dc r, đk x khác 1; -3

\(ĐKXĐ:x\ne2;x\ne4\)

\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)

\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)

\(\Rightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow-3x+8=\frac{16}{5}\left(x^2-6x+8\right)\)

\(\Rightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)

\(\Rightarrow\frac{16}{5}x^2-\frac{81}{5}x+\frac{88}{5}=0\)

Ta có \(\Delta=\frac{81^2}{5^2}-4.\frac{16}{5}.\frac{88}{5}=\frac{929}{25},\sqrt{\Delta}=\frac{\sqrt{929}}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}\)

a,\(\frac{2x-5}{3}-\frac{3x-1}{2}< \frac{3-x}{5}-\frac{2x-1}{4}\)

\(\Leftrightarrow\frac{\left(2x-5\right)20}{60}-\frac{\left(3x-1\right)30}{60}< \frac{\left(3-x\right)12}{60}-\frac{\left(2x-1\right)15}{60}\)

\(\Leftrightarrow40x-100-90x+30< 36-12x-30x+15\)

\(\Leftrightarrow40x-90x+12x+30x< 36+15+100-30\)

\(\Leftrightarrow-8x< 121\)

\(\Leftrightarrow x>-\frac{378}{25}\)

\(\hept{\begin{cases}2.\frac{1}{x}+5.\frac{1}{x+y}=2\\3.\frac{1}{x}+\frac{1}{x+y}=1,7\end{cases}}\)

Đặt \(\frac{1}{x}\)=a 

\(\frac{1}{x+y}=b\)

ta có \(\hept{\begin{cases}2a+5b=2\\3a+b=1,7\end{cases}}\)

\(\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{5}\end{cases}}\)

=> \(\frac{1}{x}=\frac{1}{2}\Rightarrow x=2\)

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