Question

Tìm số nguyên x biết

a,\(\frac{x+1}{3}=\frac{2}{6}\)

b,\(\frac{x-1}{4}=\frac{1}{-2}\)

c,\(\frac{-1}{6}=\frac{3}{2x}\)

d,\(\frac{x+1}{3}=\frac{3}{x+1}\)

e,\(\frac{4}{5}=\frac{-12}{9-x}\)

f,\(\frac{x-1}{-4}=\frac{-4}{x-1}\)

g,\(\frac{5-x}{2}=\frac{2}{5-x}\)

h,\(\frac{4-x}{-5}=\frac{-5}{4-x}\)

Giúp mình vs

Answer 1

a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)

⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)

Vậy: x=0

b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)

⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)

Vậy: x=-1

c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)

⇔\(2x=\frac{3\cdot6}{-1}=-18\)

hay x=-9

Vậy: x=-9

d) Ta có: \(\frac{x+1}{3}=\frac{3}{x+1}\)

⇔\(\left(x+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: x∈{2;-4}

e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)

⇔\(9-x=\frac{-12\cdot5}{4}=-15\)

hay x=24

Vậy: x=24

f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)

⇔\(\left(x-1\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: x∈{5;-3}

g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)

⇔\(\left(5-x\right)^2=4\)

⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

Vậy: x∈{3;7}

h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)

⇔\(\left(4-x\right)^2=25\)

⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

Vậy: x∈{-1;9}