\(\frac{x+2}{10}+\frac{x+2}{11}=\frac{x+2}{12}+\frac{x+2}{13}\)
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25 tháng 12 2015
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
=> \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
=> \(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\Rightarrow\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)
=> \(x+2=0\Rightarrow x=-2\)
23 tháng 7 2016
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}\frac{x+2}{13^{13}}\)
=> x + 2 = 0
=> x = 0 - 2
=> x = -2
31 tháng 12 2015
chuyển vế rồi phân phối, có 1/10^10+...-1/13^13 khác 0
nên x+2=0
rồi tìm x
31 tháng 12 2015
Tìm x thuộc Z, biết
( 3x+ 4) :( x-3)
x+1 là ước của 2^2+7
Trình bày ra nhé!!
LM
7 tháng 9 2016
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.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
AD
0
3 tháng 1 2016
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
PT
2
M
4 tháng 1 2016
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 \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\) => \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)
=> \(x+2=0\)
=> \(x=-2\)
4 tháng 1 2016
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 \(x+2=0\Rightarrow x=-2\)
Y
6
21 tháng 8 2016
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x +2}{12^{12}}+\frac{x+2}{13^{13}}\)
\(\Leftrightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\left(\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\right)=0\)
\(\Leftrightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
\(\Leftrightarrow\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)=0\)
Vì \(\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)\ne0\)nên \(x+2=0\Rightarrow x=-2\)
21 tháng 8 2016
<=>\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
<=>\(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}>0\)
=> \(x+2=0\)
<=>\(x=-2\)
x = -1
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