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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!!
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
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+2/10^10+x+2/11^11-x+2/12^12-x+2/13^13=0
=>(x+2).(1/10^10+1/11^11-1/12^12-1/13^13)=0
Mà 1/10^10>1/11^11>1/12^12>1/13^13
=>1/10^10+1/11^11-1/12^12-1/13^13 khác 0
=>x+2=0=>x=-2
Tick nhé
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
\(\Rightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
\(\Rightarrow\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)\ne0\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
=> \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
=> \(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\Rightarrow\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)
=> \(x+2=0\Rightarrow x=-2\)