Question

giải các pt sau:

a)7x-8= 4x+7

b)\(\frac{5x-4}{12}\)=\(\frac{16x+1}{7}\)

c) \(\frac{y+1}{y-2}\)- \(\frac{5}{y+2}\)= \(\frac{12}{y^2-4}\)+1

Answer 1

a) \(7x-8=4x+7\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

b) \(\frac{5x-4}{12}=\frac{16x+1}{7}\)

\(\Leftrightarrow35x-28=192x+12\)

\(\Leftrightarrow157x=-40\Leftrightarrow x=\frac{-40}{157}\)

Answer 2

c)\(ĐKXĐ:x\ne\pm2\)

 \(\frac{y+1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)

\(\Rightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+y^2-4}{y^2-4}\)

\(\Rightarrow\frac{y^2+3y+2-5y+10}{y^2-4}=\frac{12+y^2-4}{y^2-4}\)

\(\Rightarrow y^2-2y+12=12+y^2-4\)

\(\Rightarrow-2y=-4\Leftrightarrow y=2\left(ktm\right)\)

Vậy pt vô nghiệm

Answer 3

\(a,7x-8=4x+7\)

\(7x-4x=7+8\)

\(3x=15\Leftrightarrow5\)

\(b,\frac{5x-4}{12}=\frac{16x+1}{7}\)

\(\left(5x-4\right)7=\left(16x+1\right)12\)

\(35x-28=192x+12\)

\(35x-192x=12+28\)

\(-157x=40\Leftrightarrow x=-\frac{40}{157}\)