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Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
Do đó :
\(\frac{y+z-x}{x}=1\)\(\Rightarrow\)\(2x=y+z\)
\(\frac{z+x-y}{y}=1\)\(\Rightarrow\)\(2y=x+z\)
\(\frac{x+y-z}{z}=1\)\(\Rightarrow\)\(2z=x+y\)
Suy ra :
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{x}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)
Vậy \(P=8\)
Đề hơi sai
8:50 gửi--> 9:30 đi
=> bạn phải nhắn tin may ra có kết quả mong đợi
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow\hept{\begin{cases}y+z+1=2x\\x+z+2=2y\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)
\(A=2016x+y^{2017}+z^{2017}=2016.\frac{1}{2}+\left(\frac{5}{6}\right)^{2017}+\left(-\frac{5}{6}\right)^{2017}=1008\)
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) nha bạn!
ko hỉu thì ib
\(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)\ge9\) với x,y,z dương hay jj đó chứ? (cái này t k bt -.-) VD: x=2, y=-2,z=4
=> \(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)=\left(2-2+4\right).\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{4}\right)=1\)
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\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\frac{x+y+z}{x+y+z}=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
vì x+y+z khác 0 => \(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{xy+yz+xz}{xyz}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{\left(xy+yz+xz\right).\left(x+y+z\right)-xyz}{xzy.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\frac{x^2y+xy^2+xyz+zyx+y^2z+yz^2+x^2z+xyz+xz^2-xzy}{xyz.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x^2y+xyz\right)+\left(xy^2+y^2z\right)+\left(yz^2+xzy\right)+\left(x^2z+xz^2\right)=0\)
\(\Leftrightarrow xy.\left(x+z\right)+y^2.\left(x+z\right)+yz.\left(z+x\right)+xz.\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left[x.\left(y+z\right)+y.\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x+y\right).\left(y+z\right).\left(x+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-y\\y=-z\end{cases}\text{hoặc }x=-z}\)
\(\Rightarrow P=\left(\frac{1}{x}-\frac{1}{y}\right).\left(\frac{1}{y}+\frac{1}{z}\right).\left(\frac{1}{z}+\frac{1}{x}\right)=0\)
ps: bài này t làm cách l8, ai có cách ez hơn giải vs ak :') morongtammat
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)
\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)
Thay vào lần lượt ta có:
\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)
Tham khảo nha .
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{y+x-2}=x+y+z.\)
\(\Rightarrow\frac{x+y+z}{y+z+1+z+x+1+y+x-2}=\frac{x+y+z}{1}\) ( áp dụng tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\frac{x+y+z}{2x+2y+2z}=\frac{x+y+z}{1}\)
\(\Rightarrow2x+2y+2z=1\)
\(\Rightarrow2.\left(x+y+z\right)=1\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
Bó tay !
Ta có: \(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=x+y+z\)
TH1: \(x+y+z=0\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{x+y+z}{x+y-3+y+z+1+z+x+2}\)
\(=\frac{x+y+z}{x+y+y+z+z+x}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow x+y=\frac{1}{2}-z\)
\(y+z=\frac{1}{2}-x\)
\(z+x=\frac{1}{2}-y\)
Thay \(x+y-3=\frac{1}{2}-z-3\)
\(\Rightarrow\frac{z}{\frac{1}{2}-z+3}=\frac{1}{2}\)
\(\Rightarrow2z=\frac{1}{2}-z-3\)
\(\Rightarrow2z+z=\frac{1}{2}-3\)
\(\Rightarrow3z=-\frac{5}{2}\Rightarrow z=-\frac{5}{6}\)
Thay \(y+z+1=\frac{1}{2}-x+1\)
\(\Rightarrow\frac{x}{\frac{1}{2}-x+1}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}-x+1\)
\(\Rightarrow2x+x=\frac{1}{2}+1\)
\(\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(z+x+2=\frac{1}{2}-y+2\)
\(\Rightarrow\frac{y}{\frac{1}{2}-y+2}=\frac{1}{2}\)
\(\Rightarrow2y=\frac{1}{2}-y+2\)
\(\Rightarrow2y+y=\frac{1}{2}+2\)
\(\Rightarrow3y=\frac{5}{2}\Rightarrow y=\frac{5}{6}\)
Ta có: \(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)
\(=\left(\frac{1}{2}+\frac{5}{6}+-\frac{5}{6}-\frac{3}{2}\right)^{2019}\)
\(=\left[\left(\frac{1}{2}-\frac{3}{2}\right)+\left(-\frac{5}{6}+\frac{5}{6}\right)\right]^{2019}\)
\(=\left(-1\right)^{2019}=-1\)
TH2: x + y + z = 0
\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=0\)
\(\Rightarrow x=y=z=0\)
\(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)
\(=\left(0-\frac{3}{2}\right)^{2019}=\left(-\frac{3}{2}\right)^{2019}\)
Ah! Mk nhầm chút. TH1 là khác 0 nhé!!!!!!
b) \(\frac{x-1}{2}=\frac{2x-2}{4}\)
\(\frac{y-2}{3}=\frac{3y-6}{9}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z+3-2-6}{9}=\frac{50+3-2-6}{9}=\frac{45}{9}=5\)=>x-1=5.2=10
=>x=11
y-2=5.3=15
=>y=17
z-3=5.4=20
=>z=23
Vậy (x;y;z)=(11;17;23)
Áp dụng t/c của dãy tỉ số bằng nhau:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+x-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(vì x+y+z khác 0).Do đó x+y+z = 0.5
Thay kq này vào bài ta được:
\(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)
Tức là : \(\frac{1,5-x}{x}=\frac{2,5-y}{y}=\frac{-2,5-z}{z}=2\)
Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\)