Question

[Yumi]

Tìm \(x\) biết \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)

Answer 1

[Nguyễn Huệ Lam]

\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x +2}{12^{12}}+\frac{x+2}{13^{13}}\)

\(\Leftrightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\left(\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\right)=0\)

\(\Leftrightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

\(\Leftrightarrow\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)=0\)

Vì \(\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)\ne0\)nên \(x+2=0\Rightarrow x=-2\)

Answer 2

[Nguyễn Quang Linh]

<=>\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

<=>\(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)

Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}>0\)

=>  \(x+2=0\)

<=>\(x=-2\)