Có điều kiện gì của a,b,c không ạ?

a/ \(3a=2b;4b=3c\)

=> \(6a=4b=3c\)

=> \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{2+12-20}=\dfrac{-30}{-6}=5\)

=> \(\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

=> B

\(x\ne0;x\ne\pm5\)

PT \(\Leftrightarrow\dfrac{x+25}{2\left(x+5\right)\left(x-5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}=0\)

\(\Rightarrow x^2+25x-2x^2-20x-50+x^2-10x+25=0\)

\(\Leftrightarrow-5x-25=0\)

\(\Leftrightarrow x=-5\) (ktm)
Vậy pt vô nghiệm.

`H(x)=0`

\(\Leftrightarrow-x^2+2x-4=0\)

\(\Leftrightarrow x^2-2x+4=0\)

\(\Leftrightarrow\left(x-1\right)^2+3=0\) (vô lí)
Do đó H(x) vô nghiệm.

\(x\ne\pm2\)

\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

=> \(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow12-x^2-x-2x-2+x^2+7x-2x-14=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow x=2\) (ktm)
Vậy pt vô nghiệm.

a/ \(A=x^2y+3xy-2x^3-2x^2y+5y^4=3xy-2x^3-x^2y+5y^4\)

Thay x =1; y = 2 vào A có

\(A=6-2-2+5.16=82\)