Question

Tìm các số nguyên ,biết :

a)\(\dfrac{4+x}{x+2}=\dfrac{8}{6}\)

b)\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\)

c)\(\dfrac{x}{-2}=\dfrac{-8}{x}\)

Answer 1

Giải:

a) \(\dfrac{4+x}{x+2}=\dfrac{8}{6}\)

\(\Leftrightarrow\dfrac{4+x}{x+2}=\dfrac{4}{3}\)

\(\Leftrightarrow3\left(4+x\right)=4\left(x+2\right)\)

\(\Leftrightarrow12+3x=4x+8\)

\(\Leftrightarrow4x-3x=12-8\)

\(\Leftrightarrow x=4\)

Vậy ...

b) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\)

\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)

\(\Leftrightarrow3x+6=-4x+20\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=\dfrac{14}{7}=2\)

Vậy ...

c) \(\dfrac{x}{-2}=\dfrac{-8}{x}\)

\(\Leftrightarrow x^2=\left(-2\right)\left(-8\right)\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Vậy ...