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

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

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

Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\Rightarrow\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)

=> \(x+2=0\Rightarrow x=-2\)

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

=> x + 2 = 0

=> x       = 0 - 2

=> x       = -2

chuyển vế rồi phân phối, có 1/10^10+...-1/13^13 khác 0

nên x+2=0

rồi tìm x

Tìm x thuộc Z, biết

( 3x+ 4) :( x-3)

x+1 là ước của 2^2+7

Trình bày ra nhé!!

Đỗ Lê Tú Linh giúp mình đi, mình tick cho!!

a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)

b.

\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)

\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)

\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)

\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)

Ta có:

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

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

Mà \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\)

\(x+2=0\)

\(\Rightarrow x=-2\)

Vậy x=-2

-2

Tik mk nha...........cảm ơn

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

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

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

Vì  \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\)  => \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)

=> \(x+2=0\)

=> \(x=-2\)

Ta có:

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

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

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

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

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

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

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

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

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

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

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

=>  \(x+2=0\)

<=>\(x=-2\)

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