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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 \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=36\)
\(\text{Đặt }\frac{m}{a}=\frac{n}{b}=\frac{k}{c}=l,\text{ ta có: }\)
\(m=al,n=bl,k=cl\)
\(A=\frac{alx+bly+clz}{ax+by+cz}=\frac{l\left(ax+by+cz\right)}{ax+by+cz}=l\)
Vậy..
\(2,2.\left(x+y\right)=5.\left(y+z\right)=3.\left(x+z\right)\Leftrightarrow\frac{x+y}{5}=\frac{y+z}{2},\frac{y+z}{3}=\frac{x+z}{5}\)
\(\Leftrightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}=\frac{y+z-x-z}{6-10}=\frac{y-x}{-4}=\frac{x-y}{4}=\frac{x+y-x-z}{15-10}=\frac{y-z}{5}\)
\(\Rightarrow\frac{x-y}{4}=\frac{y-z}{5}\left(đpcm\right)\)
a)\(\frac{a^2+a+3}{a+1}=\frac{a\left(a+1\right)+3}{a+1}=\frac{a\left(a+1\right)}{a+1}+\frac{3}{a+1}=a+\frac{3}{a+1}\in Z\)
\(\Rightarrow3⋮a+1\)
\(\Rightarrow a+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow a\in\left\{0;-2;2;-4\right\}\)
b) Phần 1
\(x-2xy+y=0\)
\(\Rightarrow2x-4xy+2y=0\)
\(\Rightarrow2x-4xy+2y-1=-1\)
\(\Rightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Rightarrow\left(2x-1\right)\left(1-2y\right)=-1\)
Lập bảng xét Ư(-1)={1;-1}
Phần 2:
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
\(\Leftrightarrow\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{t+x+y}+1=\frac{t}{x+y+z}+1\)
\(\Leftrightarrow\frac{x+y+z+t}{y+z+t}=\frac{y+z+t+x}{z+t+x}=\frac{z+t+x+y}{t+x+y}=\frac{t+x+y+z}{x+y+z}\)
+)XÉt \(x+y+z+t\ne0\) suy ra \(x=y=z=t\), Khi đó \(P=1+1+1+1=4\)
+)Xét \(x+y+z+t=0\) suy ra x+y=-(z+t); y+z=-(t+x); (z+t)=-(x+y); (t+x)=-(y+z)
Khi đó \(P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Vậy P có giá trị nguyên
ĐK:y+z+t,z+t+x,t+x+z,x+z+y khác 0
x+y+t+z khác 0
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{x+y+z+t}{3\left(x+y+z+t\right)}\)
mà x+y+z+t khác 0 nên:
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{1}{3}\Rightarrow x=y=z=t\)
\(\Rightarrow P=4\left(\text{nguyên}\right).\text{Vậy: P nguyên}\)
Ta có : \(\frac{x}{x+y}>\frac{x}{x+y+z}.\)
\(\frac{y}{y+z}>\frac{y}{x+y+z}\)
\(\frac{z}{z+x}>\frac{z}{x+y+z}\)
\(\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>\)\(\frac{x+y+z}{x+y+z}=1\)
Hay \(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>1\)\(\left(1\right)\)
Lại có : \(\frac{x}{x+y}< \frac{x+z}{x+y+z}\)
\(\frac{y}{y+z}< \frac{y+x}{x+y+z}\)
\(\frac{z}{z+x}< \frac{z+y}{x+y+z}\)
\(\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< \frac{2x+2y+2z}{x+y+z}=2\)
Hay \(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)\(\left(2\right)\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)\(\left(đpcm\right)\)