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Lời giải:
Có \(M=\left ( \frac{1}{4}+\frac{3}{4^3}+...+\frac{2015}{4^{2015}} \right )-\left ( \frac{2}{4^2}+\frac{4}{4^4}+...+\frac{2016}{4^{2016}} \right )=A-B\)
Xét \(A= \frac{1}{4}+\frac{3}{4^3}+...+\frac{2015}{4^{2015}} \Rightarrow 16A=4+\frac{3}{4}+\frac{5}{4^3}+...+\frac{2015}{4^{2013}}\)
\(\Rightarrow 15A=4+2\underbrace{\left ( \frac{1}{4}+\frac{1}{4^3}+...+\frac{1}{4^{2013}} \right )}_{T}-\frac{2015}{4^{2015}}\)
Lại có \(16T=4+\frac{1}{4}+\frac{1}{4^3}+...+\frac{1}{4^{2011}}\Rightarrow 15T=4-\frac{1}{4^{2013}}\)
Do đó \(A=\frac{1}{15}\left ( 4+\frac{8}{15}-\frac{2}{15.4^{2013}}-\frac{2015}{4^{2015}} \right )\)
Thực hiện tương tự, suy ra
\(B=\frac{1}{15}\left ( 2+\frac{2}{15}-\frac{2}{15.4^{2014}}-\frac{2016}{4^{2016}} \right )\)
\(\Rightarrow M=A-B=\frac{1}{15}\left ( \frac{12}{5}-\frac{90692}{15.4^{2014}} \right )<\frac{1}{15}.\frac{12}{5}=\frac{4}{25}\)
Ta có đpcm
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{\left(5x+1\right)\left(5x+6\right)}=\dfrac{2015}{2016}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5x+1}-\dfrac{1}{5x+6}=\dfrac{2015}{2016}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{5x+6}=\dfrac{2015}{2016}\)
\(\Rightarrow\dfrac{5x+5}{5x+6}=\dfrac{2015}{2016}\)
\(\Rightarrow\left(5x+5\right).2016=\left(5x+6\right).2015\)
\(\Rightarrow10080x+10080=10075+12090\)
\(\Rightarrow5x=2010\)
\(\Rightarrow x=402\)
Vậy x = 402
Ta có:
\(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)
\(\Leftrightarrow xy\ge1\)
Theo đề bài thì
\(\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{y^4+x^2+2yx^2}\le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}\le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
theo bđt cauchy schwarz ta có
\(\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2\sqrt{y^3z^2}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2\sqrt{z^3y^2}}=\dfrac{1}{zy}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}+\dfrac{\dfrac{1}{y^2}+\dfrac{1}{z^2}}{2}+\dfrac{\dfrac{1}{z^2}+\dfrac{1}{x^2}}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)\(\Rightarrow dpcm\)