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Đặt: \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k\)
\(\Rightarrow x=k\)
\(y=2k\)
\(z=3k\)
Thay x = k , y = 2k , z = 3k vào biểu thức cần cm ,ta đc:
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=\left(k+2k+3k\right)\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)
\(=6k.\left(\frac{1}{k}+\frac{2}{k}+\frac{3}{k}\right)\)
\(=6k.\frac{6}{k}\)
\(=\frac{36k}{k}=36\)
=.= hok tốt!!
Đặt \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k\)
Do đó \(x=k;y=2k;z=3k\)
Thay \(x=k;y=2k;z=3k\)vào \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)\)ta có
\(\left(k+2k+3k\right).\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)
\(=6k.\left(\frac{6}{6k}+\frac{12}{6k}+\frac{18}{6k}\right)\)
\(=6k.\frac{6+12+18}{6k}\)
\(=\frac{6k.\left(6+12+18\right)}{6k}\)
\(=36\)
Do đó \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=36\)
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x-y}{2-3}=\frac{y-z}{3-4}=\frac{x-z}{2-4}\) (T/c dãy tỷ số bằng nhau)
\(\Rightarrow\frac{x-z}{-2}=-\left(x-y\right)\left(1\right)\Rightarrow\frac{\left(x-z\right)^3}{-8}=-\left(x-y\right)^3=-\left(x-y\right)^2\left(x-y\right)\left(2\right)\)
\(\Rightarrow\frac{x-z}{-2}=-\left(y-z\right)\left(3\right)\)
Từ (1) và (3) \(\Rightarrow\left(x-y\right)=\left(y-z\right)\) Thay vào (2)
\(\Rightarrow\frac{\left(x-z\right)^3}{-8}=-\left(x-y\right)^2\left(y-z\right)\Rightarrow\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\left(dpcm\right)\)
\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
\(A=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{y+z}{z}\)
Do \(x-y-z=0\)
\(\Rightarrow x-z=y;y-x=-z;y+z=x\)
Khi đó \(A=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
Vậy A=-1
\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)
\(=\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{1+yz+y}\)
\(=\frac{1}{xy+x+1}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz}{xy\cdot yz+xyz+yz}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz}{yz+y+1}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz+y+1}{yz+y+1}\)
\(=1\)
6x=3y=2z nên 6x/6=3y/6=2z/6
=>x/1=y/2=z/3=k
=>x=k; y=2k; z=3k
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\right)^2\)
\(=\left(k+2k+3k\right)\cdot\left(\dfrac{1}{k}+\dfrac{4}{2k}+\dfrac{9}{3k}\right)^2\)
\(=6k\cdot\left(\dfrac{1}{k}+\dfrac{2}{k}+\dfrac{3}{k}\right)^2=6k\cdot\dfrac{36}{k^2}=\dfrac{6}{k}\)
a, \(x:y:z=2:3:4\&x+y+z=365\)
\(x:y:z=2:3:4\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tích chất dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{365}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{365}{9}\\\dfrac{y}{3}=\dfrac{365}{9}\\\dfrac{z}{4}=\dfrac{365}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{730}{9}\\y=\dfrac{365}{3}\\z=\dfrac{1460}{9}\end{matrix}\right.\)
b:\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\\dfrac{7}{2}+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)
c: =>1/2x-5=0 và y^2-1/4=0
=>\(\left\{{}\begin{matrix}x=10\\y\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\end{matrix}\right.\)
d: =>x=0 và y-1/10=0
=>x=0 và y=1/10
Ta có:\(x:y:z=1:2:3\Rightarrow x=\frac{y}{2}=\frac{z}{3}\).Đặt \(x=\frac{y}{2}=\frac{z}{3}=k\)
\(\Rightarrow\hept{\begin{cases}x=k\\y=2k\\z=3k\end{cases}}\)\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=\left(k+2k+3k\right)\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)
\(=6k.\left(\frac{1}{k}+\frac{2}{k}+\frac{3}{k}\right)=6k.\frac{6}{k}=36\)
\(\Rightarrowđpcm\)