Question

tìm số nguyên x biết :

\(a,\frac{x}{5}=\frac{2}{5}\) \(b,\frac{3}{8}=\frac{6}{x}\) \(c,\frac{1}{9}=\frac{x}{27}\)

\(d,\frac{4}{x}=\frac{8}{6}\) \(c,\frac{3}{x-5}=\frac{-4}{x+2}\) \(g,\frac{x}{-2}+\frac{-8}{x}\)

giúp mk với !!!!!!

Answer 1

a) Ta có: \(\frac{x}{5}=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{2\cdot5}{5}=2\)

Vậy: x=2

b) Ta có: \(\frac{3}{8}=\frac{6}{x}\)

\(\Leftrightarrow x=\frac{6\cdot8}{3}=\frac{48}{3}=16\)

Vậy: x=16

c) Ta có: \(\frac{1}{9}=\frac{x}{27}\)

\(\Leftrightarrow x=\frac{1\cdot27}{9}=\frac{27}{9}=3\)

Vậy: x=3

d) Ta có: \(\frac{4}{x}=\frac{8}{6}\)

\(\Leftrightarrow x=\frac{4\cdot6}{8}=\frac{24}{8}=3\)

Vậy: x=3

e) Ta có: \(\frac{3}{x-5}=\frac{-4}{x+2}\)

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

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

\(\Leftrightarrow3x+6+4x-20=0\)

\(\Leftrightarrow7x-14=0\)

\(\Leftrightarrow7x=14\)

hay x=2

Vậy: x=2

g) Sửa đề: \(\frac{x}{-2}=\frac{-8}{x}\)

Ta có: \(\frac{x}{-2}=\frac{-8}{x}\)

\(\Leftrightarrow x^2=\left(-8\right)\cdot\left(-2\right)=16\)

hay x∈{4;-4}

Vậy: x∈{4;-4}

Answer 2

a) \(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\cdot5}{5}=2\)

b)\(\frac{3}{8}=\frac{6}{x}\Rightarrow x=\frac{6\cdot8}{3}=16\)

c)\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{27}{9}=3\)

d) \(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4\cdot6}{8}=3\)

e)

\(\frac{3}{x-5}=-\frac{4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+6=-4x+20\\ \Leftrightarrow3x+6+4x-20=0\\ \Leftrightarrow7x-14=0\\ \Leftrightarrow x=2\)

\(g,\frac{x}{-2}=-\frac{8}{x}\Rightarrow x^2=16\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Vậy có 2 giá trị của x là 4 ; -4