Question

Tìm x\(\frac{x-1}{9}=\frac{8}{3}\)
b) \(\frac{x}{4}=\frac{18}{x+1}\)
c) \(\frac{-x}{4}=\frac{-9}{x}\)

Answer 1

\(\frac{x-1}{9}=\frac{8}{3}\)

\(\left(x-1\right)\cdot3=8\cdot9\)

\(\left(x-1\right)\cdot3=72\)

\(x-1=\frac{72}{3}\)

\(x-1=24\)

\(x=24+1\)

\(x=25\)

\(\frac{x}{4}=\frac{18}{x+1}\)

\(x\cdot\left(x+1\right)=18\cdot4\)

\(x\left(x+1\right)=72\)

\(x\left(x+1\right)=8\cdot9\)

\(x=8\)

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\frac{x}{-4}=\frac{-9}{x}\)

\(x\cdot x=\left(-9\right)\cdot\left(-4\right)\)

\(x^2=36\)

\(x^2=\left(-6\right)^2\)hoặc \(x^2=6^2\)

\(x=-6\)  hoặc\(x=6\)

Answer 2

a)x-1/9=24/9       =>  x-1=24 =>x=23

b)x(x+1)=18*4   =>x=8

c)-x:4=-9:x     =>-1.x²=-1.36    =>x=6

k mik nha!