Question

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

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

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

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

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

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

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

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

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

=>  \(x+2=0\)

<=>\(x=-2\)

Answer 3

x + 2/10¹⁰ + x + 2/11¹¹ = x + 2/12¹² + x + 2/13¹³

=> x + 2/10¹⁰ + x + 2/11¹¹ - x + 2/12¹² - x + 2/13¹³ = 0

=> (x + 2).(1/10¹⁰ + 1/11¹¹ - 1/12¹² - 1/13¹³) = 0

Vì 1/10¹⁰ > 1/12¹²; 1/11¹¹ > 1/13¹³

=> 1/10¹⁰ + 1/11¹¹ - 1/12¹² - 1/13¹³ khác 0

=> x + 2 = 0

=> x = -2

Answer 4

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

Thì khi đó x = -2 để (-2)+2 = 0 [ 0 chia với bất kì số nào cũng bằng 0]

Vậy x = -2

Answer 5

Nguyễn Huệ Lam viết sai rồi

Answer 6

yumi hả mk nè