Ta có \(\left(-\frac{1}{25}\right)5=\left(-\frac{1}{5}\right)^{2.5}=\left(-\frac{1}{5}\right)^{10}>\left(-\frac{1}{5}\right)^9\)
Ta có \(\left(-\frac{1}{25}\right)5=\left(-\frac{1}{5}\right)^{2.5}=\left(-\frac{1}{5}\right)^{10}>\left(-\frac{1}{5}\right)^9\)
so sánh:
\(\left[-\frac{1}{5}\right]^9và\left[-\frac{1}{25}\right]^5\)
Cho A = \(\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)và B = \(\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
So sánh A và B
Cho A = \(\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\); B = \(\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
So sánh A và B
So sánh với 3
\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+....+\left(1+\frac{2}{n^2+3n}\right)\)
A=\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)\)
So sánh A với 3
Cho \(A=\frac{\left(3\frac{2}{15}+\frac{1}{15}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)
\(B=\frac{1;2:\left(1\frac{1}{5}:1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
So sánh A và B
\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(100-\frac{2}{25}\right)}{\left(6\frac{5}{9}-3\frac{1}{4}\right)}+\left(1,2.0,5\right):\frac{3}{5}=\)
\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(100-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{27}}+\left(1,2.0,5\right):\frac{3}{5}\)
So sánh:
60^5 và 15^10
\(\left(\frac{1}{20}\right)^7và\left(\frac{1}{5}\right)^9\)