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\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2015^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\right)\)
\(=1-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\right)>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=1-\left(1-\frac{1}{2015}\right)=1-\frac{2014}{2015}=\frac{1}{2015}\)
=> \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2015^2}>\frac{1}{2015}\left(\text{đpcm}\right)\)
\(4.\left(-\frac{1}{2}\right)^3-2.\left(-\frac{1}{2}\right)^2-3.\left(-\frac{1}{2}\right)^1\)
\(=4.\left(-\frac{1}{2}.\left(-\frac{1}{2}\right).\left(-\frac{1}{2}\right)\right)-2.\left(-\frac{1}{2}.\left(-\frac{1}{2}\right)\right)-3.\left(-\frac{1}{2}\right)\)( sửa ngoặc vuông giúp mk )
\(=4.\left(-\frac{1}{8}\right)-2.\left(\frac{1}{4}\right)-3.\left(-\frac{1}{2}\right)\)
\(=-\frac{1}{2}-\frac{1}{2}+\frac{3}{2}\)
\(=1+\frac{3}{2}\)
\(=\frac{5}{2}\)
\(4\times\left(-\frac{1}{2}\right)^3-2\times\left(-\frac{1}{2}\right)^2-3\times\left(-\frac{1}{2}\right)^1\)
\(=4\times\left(-\frac{1}{2}\right)^3-2\times\left(-\frac{1}{2}\right)^2-3\times\left(-\frac{1}{2}\right)\)
\(=4\times\left(-\frac{1}{8}\right)-2\times\frac{1}{4}-3\times\left(-\frac{1}{2}\right)\)
\(=-\frac{1}{2}-\frac{1}{2}+\frac{3}{2}\)
\(=1+\frac{3}{2}\)
\(=\frac{2}{2}+\frac{3}{2}\)
\(=\frac{5}{2}\)
\(4.\left(-\frac{1}{2}\right)^3-3.\left(-\frac{1}{2}\right)^2-3.\left(-\frac{1}{2}\right)^1=4.-\frac{1}{8}-3.\frac{1}{4}-3.-\frac{1}{2}\)
\(=-\frac{1}{2}-\frac{3}{4}-\left(-\frac{3}{2}\right)\)
\(=-\frac{1}{2}-\frac{3}{4}+\frac{3}{2}\)
\(=\left(-\frac{1}{2}+\frac{3}{2}\right)-\frac{3}{4}\)
\(=1-\frac{3}{4}\)
\(=\frac{1}{4}\)
Hok tốt nha^^
a: \(\left(-\dfrac{1}{16}\right)^{100}=\left(\dfrac{1}{16}\right)^{100}=\left(-\dfrac{1}{2}\right)^{400}\)
\(\left(-\dfrac{1}{2}\right)^{500}=\left(-\dfrac{1}{2}\right)^{500}\)
mà \(400< 500\)
nên \(\left(-\dfrac{1}{16}\right)^{100}< \left(-\dfrac{1}{2}\right)^{500}\)