Tìm x\(\left(\frac{1}{125}\right)^x<\frac{1}{25}\)
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11 tháng 8 2019
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)
\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1}{2009}\)
19 tháng 10 2019
\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(0\)
1 tháng 8 2019
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
1 tháng 9 2016
ĐKXĐ : \(x\ge0,x\ne25,x\ne9\)
a) \(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-\left(x-25\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\left(\frac{-\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=-\frac{5}{\sqrt{x}+5}:\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}=\frac{-5}{\sqrt{x}+5}.\left(\frac{-\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\right)=\frac{5}{\sqrt{x}+3}\)
b) \(A< 1\Rightarrow\frac{5}{\sqrt{x}+3}< 1\Rightarrow\sqrt{x}+3>5\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
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