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\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)
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 \(P=\left(2.\frac{1}{2}\right)^{2019}=1\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)
\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{z}=0\\\frac{1}{y}+\frac{1}{z}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{x}=\frac{1}{-z}\\\frac{1}{y}=\frac{1}{-z}\end{cases}\Leftrightarrow}\frac{1}{x}=\frac{1}{y}=\frac{1}{-z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Leftrightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\)
\(\Leftrightarrow z=\frac{-1}{2}\)
\(x=y=\frac{1}{2}\)
\(\Rightarrow C=\left(x+2y+z\right)^{2021}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2021}=1\)
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 :
\(x=y=\frac{1}{2};z=-\frac{1}{2}\)
\(\Rightarrow P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2020}=1\)
Bài \(1a.\) Tìm \(x,y,z\) biết \(x^2+4y^2=2xy+1\) \(\left(1\right)\) và \(z^2=2xy-1\) \(\left(2\right)\)
Cộng \(\left(1\right)\) và \(\left(2\right)\) vế theo vế, ta được:
\(x^2+4y^2+z^2=4xy\)
\(\Leftrightarrow\) \(x^2-4xy+4y^2+z^2=0\)
\(\Leftrightarrow\) \(\left(x-2y\right)^2+z^2=0\)
Do \(\left(x-2y\right)^2\ge0\) và \(z^2\ge0\) với mọi \(x,y,z\)
nên để thỏa mãn đẳng thức trên thì phải đồng thời xảy ra \(\left(x-2y\right)^2=0\) và \(z^2=0\)
\(\Leftrightarrow\) \(^{x-2y=0}_{z^2=0}\) \(\Leftrightarrow\) \(^{x=2y}_{z=0}\)
Từ \(\left(2\right)\), với chú ý rằng \(x=2y\) và \(z=0\), ta suy ra:
\(2xy-1=0\) \(\Leftrightarrow\) \(2.\left(2y\right).y-1=0\) \(\Leftrightarrow\) \(4y^2-1=0\) \(\Leftrightarrow\) \(y^2=\frac{1}{4}\) \(\Leftrightarrow\) \(y=\frac{1}{2}\) hoặc \(y=-\frac{1}{2}\)
\(\text{*)}\) Với \(y=\frac{1}{2}\) kết hợp với \(z=0\) \(\left(cmt\right)\) thì \(\left(2\right)\) \(\Rightarrow\) \(2.x.\frac{1}{2}-1=0\) \(\Leftrightarrow\) \(x=1\)
\(\text{*)}\) Tương tự với trường hợp \(y=-\frac{1}{2}\), ta cũng dễ dàng suy ra được \(x=-1\)
Vậy, các cặp số \(x,y,z\) cần tìm là \(\left(x;y;z\right)=\left\{\left(1;\frac{1}{2};0\right),\left(-1;-\frac{1}{2};0\right)\right\}\)
\(b.\) Vì \(x+y+z=1\) nên \(\left(x+y+z\right)^2=1\)
\(\Leftrightarrow\) \(x^2+y^2+z^2+2\left(xy+yz+xz\right)=1\) \(\left(3\right)\)
Mặt khác, ta lại có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) \(\Rightarrow\) \(xy+yz+xz=0\) \(\left(4\right)\) (do \(xyz\ne0\))
Do đó, từ \(\left(3\right)\) và \(\left(4\right)\) \(\Rightarrow\) \(x^2+y^2+z^2=1\)
Vậy, \(B=1\)
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
\(\left(\frac{1}{x}+\frac{1}{y}\right)^2=\left(2-\frac{1}{z}\right)^2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}-\frac{1}{z^2}=4-\frac{4}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\Rightarrow\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}=-8\)
\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}-2=0\\\frac{1}{y}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\end{matrix}\right.\) \(\frac{1}{z}=2-\left(\frac{1}{x}+\frac{1}{y}\right)=-2\Rightarrow z=\frac{-1}{2}\)
Vậy \(P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2012}=1^{2012}=1\)
\(x+y+z=7\Rightarrow z=7-x-y\Rightarrow xy+z-6=xy+7-x-y-6=xy-x-y+1\)
\(=\left(x-1\right)\left(y-1\right)\)
Tương tự: \(yz+x-6=\left(y-1\right)\left(z-1\right);zx+y-6=\left(z-1\right)\left(x-1\right)\)
Viết lại: \(H=\frac{1}{\left(x-1\right)\left(y-1\right)}+\frac{1}{\left(y-1\right)\left(z-1\right)}+\frac{1}{\left(z-1\right)\left(x-1\right)}\)
\(=\frac{x-1+y-1+z-1}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=\frac{x+y+z-3}{xyz-\left(xy+yz+zx\right)+x+y+z-1}\)
\(=\frac{7-3}{3-13+7-1}=-1\)(Từ gt tính được \(xy+yz+zx=13\))
Ta có :
\(xy+yz+zx\)= \(\frac{\left(x+y+z\right)^2-x^2-y^2-z^2}{2}\)= \(\frac{7^2-23}{2}\)= \(13\)
Ta lại có :
\(xy+z-6=xy+z+1-x-y-z\)= \(\left(x-1\right)\left(y-1\right)\)
\(\Rightarrow A=\)\(\frac{1}{\left(x-1\right)\left(y-1\right)}\)\(+\)\(\frac{1}{\left(y-1\right)\left(z-1\right)}\)\(+\)\(\frac{1}{\left(z-1\right)\left(x-1\right)}\)
\(=\)\(\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}\)
\(=-1\)
=56 phân số bất đồng trị của a+b
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 \(P=\left(x+2y+z\right)^{2019}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2019}=1^{2019}=1\)