a ) Cho biết : \(x+y+z=2020\)
và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{202}\)
Tính M = \(\frac{x+y}{z}=\frac{x+z}{y}=\frac{y+z}{x}\)
b ) Cho \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(CMR:a< \frac{504}{1009}\)
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M = x+y/z + x+z/y + y+z/x
M = x+y+z/z + x+y+z/y + x+y+z/x - z/z - y/y - x/x
M = (x+y+z).(1/z + 1/y + 1/x) - 1 - 1 - 1
M = 2020.1/202 - 3
M = 10 - 3 = 7
1
Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
Lời giải:
Từ điều kiện đề bài suy ra $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Leftrightarrow \frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)\left[\frac{1}{xy}+\frac{1}{z(x+y+z)}\right]=0$
$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Rightarrow (x+y)(y+z)(x+z)=0$
Do đó: $M=\frac{x+y}{z}.\frac{x+z}{y}.\frac{y+z}{x}=\frac{(x+y)(y+z)(x+z)}{xyz}=\frac{0}{xyz}=0$
\(\frac{x2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự ta có:
\(\frac{y^2}{x+z}+y=\frac{y\left(x+y+z\right)}{x+z};\frac{z^2}{x+y}+z=\frac{z\left(x+y+z\right)}{x+y}\)
Cộng vế theo vế ta có:
\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+x+y+z=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+2020=2020\)
E ms bt bài này thôi ạ
Ta có : M = \(\frac{x+y}{z}+\frac{x+z}{y}=\frac{y+z}{x}\)
\(\Rightarrow M+3=\left(\frac{x+y}{z}+1\right)+\left(\frac{x+z}{y}+1\right)+\left(\frac{y+z}{x}+1\right)\)
\(\Rightarrow M+3=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)
\(\Rightarrow M+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow M+3=2020.\frac{1}{202}\)
=> M + 3 = 10
=> M = 7
Vậy M = 7
b) Ta có : \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(=\frac{2}{3.3}+\frac{2}{5.5}+\frac{2}{7.7}+...+\frac{2}{2017.2017}\)
\(< \frac{2}{\left(3+1\right)\left(3-1\right)}+\frac{2}{\left(5-1\right)\left(5+1\right)}+\frac{2}{\left(7-1\right)\left(7+1\right)}+...+\frac{2}{\left(2017-1\right)\left(2016-1\right)}\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}\)
\(=\frac{1008}{2018}=\frac{504}{1009}\)
=> \(A< \frac{504}{1009}\left(\text{ĐPCM}\right)\)