Cho x,y eZ , x,y khác -1 t/m \(\dfrac{x^4-1}{y+1}+\dfrac{y^4-4}{x+1}\) là số nguyên.
C/m \(x^4y^{44}-1⋮y+1\)
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29 tháng 11 2021
\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân VTV
\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)
29 tháng 11 2021
\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)
14 tháng 3 2017
Ta có :\(\dfrac{x}{y+z}=\dfrac{123-\left(y+z\right)}{y+z}\)
\(\dfrac{y}{x+z}=\dfrac{123-\left(x+z\right)}{x+z}\)
\(\dfrac{z}{y+x}=\dfrac{123-\left(y+x\right)}{y+x}\)
\(\Rightarrow P=\dfrac{123-\left(y+z\right)}{y+z}+\dfrac{123-\left(z+x\right)}{z+x}+\dfrac{123-\left(y+x\right)}{y+x}\)\(\Rightarrow P=123\left(\dfrac{1}{y+z}+\dfrac{1}{x+y}+\dfrac{1}{z+x}\right)-3\)
\(\Rightarrow P=123.\dfrac{1}{45}-3\)
\(\Rightarrow P=-\dfrac{4}{15}\)
NT
22 tháng 12 2017
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
T
2
13 tháng 4 2022
\(x+y\ge2\sqrt{xy}\)
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)
Do đó: \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)>=4\)
HN
30 tháng 6 2017
Ta có:
\(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)
\(\Leftrightarrow xy\ge1\)
Theo đề bài thì
\(\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{y^4+x^2+2yx^2}\le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}\le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
10 tháng 3 2018
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)