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18 tháng 3 2019

Đặt: \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)

\(\Leftrightarrow a^2+b^2+2bc+2ac+2c^2=0\)

\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)

\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\Leftrightarrow a=b=-c\Leftrightarrow x=y=-z\)

Giải tiếp nhé

NV
18 tháng 3 2019

\(\left(\frac{1}{x}+\frac{1}{y}\right)^2=\left(2-\frac{1}{z}\right)^2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}-\frac{1}{z^2}=4-\frac{4}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\Rightarrow\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}=-8\)

\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}-2=0\\\frac{1}{y}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\end{matrix}\right.\) \(\frac{1}{z}=2-\left(\frac{1}{x}+\frac{1}{y}\right)=-2\Rightarrow z=\frac{-1}{2}\)

Vậy \(P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2012}=1^{2012}=1\)

\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)

Khai triển cả 2 vế ta được \(\left(\frac{1}{y}+\frac{1}{z}\right)^2+\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)

=>\(\hept{\begin{cases}\frac{1}{y}+\frac{1}{z}=0\\\frac{1}{x}+\frac{1}{z}=0\end{cases}}\)=>\(\frac{1}{x}=\frac{1}{y}\Rightarrow x=y\)

=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{x}+\frac{1}{z}=2\Rightarrow\frac{4}{x^2}+\frac{4}{xz}+\frac{1}{z^2}=4\)(1)

\(\frac{2}{xy}-\frac{1}{z^2}=\frac{2}{x^2}-\frac{1}{z^2}=4\)(2)

Từ (1) và (2) suy ra

\(\frac{2}{x^2}+\frac{4}{xz}+\frac{2}{z^2}=0\Rightarrow\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}=0\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)\(\Rightarrow\frac{1}{x}+\frac{1}{z}=0\Rightarrow x=y=-z\)

=> \(P=\left(x+2y+z\right)^{2019}=\left(2y\right)^{2019}\)

à thêm cái này nữa. Sorry viết thiếu

Vì x=y=-z\(\Rightarrow\frac{2}{x}-\frac{1}{x}=2\Rightarrow\frac{1}{x}=2\Rightarrow x=\frac{1}{2}.\)

lúc đó  \(P=\left(2.\frac{1}{2}\right)^{2019}=1\)

9 tháng 3 2021

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)

\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{z}=0\\\frac{1}{y}+\frac{1}{z}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{x}=\frac{1}{-z}\\\frac{1}{y}=\frac{1}{-z}\end{cases}\Leftrightarrow}\frac{1}{x}=\frac{1}{y}=\frac{1}{-z}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Leftrightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\)

\(\Leftrightarrow z=\frac{-1}{2}\)

\(x=y=\frac{1}{2}\)

\(\Rightarrow C=\left(x+2y+z\right)^{2021}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2021}=1\)

9 tháng 3 2021

Ta có:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)

\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\\\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=-\frac{1}{z}\\\frac{1}{y}=-\frac{1}{z}\end{cases}}}\)

\(\Leftrightarrow x=y=-z\)

Thay vào \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)ta được :

\(x=y=\frac{1}{2};z=-\frac{1}{2}\)

\(\Rightarrow P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2020}=1\)

5 tháng 9 2020

=56 phân số bất đồng trị của a+b

5 tháng 9 2020

chắc câu này a đăng lên cho vui :vv

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2< =>\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2=4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\left(\frac{2}{xy}-\frac{1}{z^2}\right)+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}+4=4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{xy}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4-4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)

\(< =>\left(\frac{1}{x^2}+\frac{2}{zx}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(< =>\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0< =>\frac{1}{x}=\frac{1}{y}=-\frac{1}{z}\)

\(< =>x=y=-z\)Thế vào giả thiết ta được : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(< =>\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2< =>\frac{-1}{z}+\frac{-1}{z}+\frac{1}{z}=2\)

\(< =>\frac{-1-1+1}{z}=2< =>2z=-1< =>z=-\frac{1}{2}\)

Suy ra \(x=y=-z=-\left(-\frac{1}{2}\right)=\frac{1}{2}< =>\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{cases}}\)

Nên \(P=\left(x+2y+z\right)^{2019}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2019}=1^{2019}=1\)

1 tháng 3 2020

áp dụng tính chất của dãy tỉ số bằng nhau ta có:\(\frac{ }{ }\)

y+z-x/x=z+x-y/y=x+y-z/z

=y+z-x+z+x-y+x+y-z/x+y+z

=(y-y)+(z-z)-(x-x)+z+x+y/x+y+z

=0+0+0+x+y+z/x+y+z=1

\(\Leftrightarrow\)x=y=z (*)

thay (*) vào B ta có:

B=(1+x/x)(1+x/x)(1+x/x)

  =2.2.2=8

21 tháng 8 2020

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(...=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)( vì x + y + z \(\ne\)0 )

\(\Rightarrow\hept{\begin{cases}\frac{y+z-x}{x}=1\\\frac{z+x-y}{y}=1\\\frac{x+y-z}{z}=1\end{cases}}\Rightarrow\hept{\begin{cases}y+z-x=x\\z+x-y=y\\x+y-z=z\end{cases}}\Rightarrow\hept{\begin{cases}y+z=2x\\z+x=2y\\x+y=2z\end{cases}}\Rightarrow x=y=z\)

Thế x = y = z vào B ta được :

\(B=\left(1+\frac{y}{y}\right)\left(1+\frac{x}{x}\right)\left(1+\frac{z}{z}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2\cdot2\cdot2=8\)

NV
20 tháng 6 2019

\(\left(\frac{1}{x}+\frac{1}{y}\right)^2=\left(2-\frac{1}{z}\right)^2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4-\frac{4}{z}+\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=\frac{2}{xy}-\frac{1}{z^2}-\frac{4}{z}+\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}=4\left(\frac{1}{x}+\frac{1}{y}-2\right)\)

\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\)

\(\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\Rightarrow x=y=\frac{1}{2}\Rightarrow z=-\frac{1}{2}\)

\(\Rightarrow\left(x+2y+z\right)^{2012}=1^{2012}\)