\(1,\frac{7x-3}{x-1}=\frac{2}{3}\)    ĐKXĐ : \(x\ne1\)

\(\Leftrightarrow\frac{3\left(7x-3\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Rightarrow21x-2x=9-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)(TM)

kl :....

\(3,\frac{1}{x-2}+3=\frac{x-3}{2-x}\)   ĐKXĐ : \(x\ne2\)

\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)

\(\Leftrightarrow1+3x-6=3-x\)

\(\Leftrightarrow3x+x=-1+6-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=2\)(TM)

KL : ....

đề bài ko rõ nên mik ko làm đc( H AE là gì?)

H thuộc AE

a) \(x^2-3x+3=0\)

\(\Leftrightarrow x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{3}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{-3}{4}\)

Vô lí => Phương trình vô nghiệm

\(\Rightarrow x\in\varnothing\)

b) \(x-\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow x-x+2-x+2=0\)

\(\Leftrightarrow-x=-4\)

\(\Leftrightarrow x=4\)

c) \(\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}}\)

Vậy x = {1;2} 

d) \(x^2-2x-x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy x = {1;2}

ta có : \(\frac{x}{7}=\frac{y}{13}\)

ADTCDTSBN : 

\(\frac{x}{7}=\frac{y}{13}=\frac{x+y}{7+13}=\frac{40}{20}=2\)

\(\Rightarrow\frac{x}{7}=2\Rightarrow x=14\)

\(\frac{y}{13}=2\Leftrightarrow y=26\)

KL : ... 

kb nhoa 

Viết sai chữ Thanks kìa