Question

Tìm x biết :

1) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

2) \(\left(x-2,5\right):1\frac{1}{2}=x:2\)

3) \(\left(\frac{1}{3}+x\right):\left(x+\frac{3}{4}\right)=\frac{7}{9}\)

4) \(0,81:\frac{x}{2}=\frac{16}{x^4}:\left(-0,9\right)\)

giúp mình nhé 

Answer 1

1)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)

\(\Leftrightarrow x=-1\)

Answer 2

\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\\ \Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\\ \Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Gọi \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\)là A

Vì A ≠0

⇒\(x+1=0\Rightarrow x=-1\)

Vậy \(x=-1\)