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\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
\(\Rightarrow A=0\)
Ta có : \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
Khi đó : \(3x^{2018}=27^{673}=\left(3^3\right)^{673}=3^{2019}\)
\(\Leftrightarrow x^{2018}=3^{2018}\)
\(\Leftrightarrow\orbr{\begin{cases}x=y=z=3\\x=y=z=-3\end{cases}}\)
Đến đây tự tính A nha!
Sửa đề : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2019}\)
Thay \(2019=x+y+z\)ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{y}{xy}+\frac{x}{xy}=\frac{z}{z\left(x+y+z\right)}-\frac{x+y+z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\left(x+y\right)\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)+xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)
( mình chỉ xét 1 t/h, các t/h còn lại hoàn toàn tương tự )
TH1 : \(x+y=0\)
\(\Leftrightarrow x=-y\)(1)
Thay (1) vào A ta có :
\(A=\frac{1}{-y^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}\)
\(A=\frac{1}{z^{2019}}\)
Mặt khác : \(x+y+z=2019\)
Thay (1) vào đẳng thức trên ta được : \(-y+y+z=2019\)
\(\Leftrightarrow z=2019\)
Thay z vào A ta được : \(A=\frac{1}{2019^{2019}}\)
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)
Từ \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)
Suy ra: x=y=z
\(\Rightarrow3x^{2018}=3y^{2018}=3z^{2018}=27^{673}=3^{2019}\)
\(\Leftrightarrow x^{2018}=y^{2018}=z^{2018}=3^{2018}\)
\(\Rightarrow x,y,z=3\)
Dễ tính A
Bài 2:
\(x^3+y^3+z^3-3xyz=0\)
<=> \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
<=> \(\left\{{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{matrix}\right.\)
Ta có \(a^2+b^2+c^2\ge ab+bc+ca\)
Áp dụng => \(x^2+y^2+z^2\ge xy+yz+zx\)
Dấu "=" xảy ra <=> x = y = z (vô lí do x,y,z đôi 1 khác nhau)
=> x + y + z =0
=> \(\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\z+x=-y\end{matrix}\right.\)
Thay vào P = -16 - 3 + 2019 = 2000
Bài 1:
Ta có: \(x^2+y^2+5x^2y^2+60=37xy\)
\(\Leftrightarrow x^2+y^2-2xy+60=35xy-5x^2y^2\)
\(\Leftrightarrow\left(x-y\right)^2+60=5\left(7xy-x^2y^2\right)\)
\(\Leftrightarrow\left(x-y\right)^2+60=\frac{5\cdot49}{4}-\frac{5}{4}\left(2xy-7\right)^2\)
\(\Leftrightarrow\left[2\left(x-y\right)\right]^2+5\left(2xy-7\right)^2=5\cdot49-60\cdot4=5\)
mà \(x,y\in Z\) và \(2xy-7\ne0\); \(5\left(2xy-7\right)^2\ge5\)
nên \(\left[2\left(x-y\right)\right]^2=0\)
\(\Leftrightarrow x=y\)
|(2xy-7)|=1
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-7=-1\\2x^2-7=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2=6\\2x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=3\left(loại\right)\\x^2=4\end{matrix}\right.\)
\(\Leftrightarrow x=\pm2\)
Vậy: (x,y)=(\(\pm2;\pm2\))
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)=> \(\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)
=> (x+y+z)(xy+yz+zx) = xyz
=> \(x^2y+xy^2+y^2z+yz^2+zx^2+z^2x+2xyz=0\)
=> (x+y)(y+z)(z+x) = 0
=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: x = -y
=> \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{\left(-y\right)^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{z^{2019}}\)
=> \(\frac{1}{x^{2019}+y^{2019}+z^{2019}}=\frac{1}{\left(-y\right)^{2019}+y^{2019}+z^{2019}}=\frac{1}{z^{2019}}\)
=> ĐPCM
Tương tự với TH2 và TH3
Ta có: \(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)^3-3.\left(x+y\right).z.\left(x+y+z\right)-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right).\left[\left(x+y+z\right)^2-3.\left(x+y\right).z\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2zx-3xz-3yz-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2-xz-yz-xy\right)=0\)
+ \(x+y+z=0\)\(\Rightarrow\)\(C=\frac{x^{2019}+y^{2019}+z^{2019}}{0}\)( Loại )
+ \(x^2+y^2+z^2-xz-yz-xy=0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xz-2yz-2xy=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\)\(x=y=z\)
\(\Rightarrow\)\(C=\frac{x^{2019}+x^{2019}+x^{2019}}{\left(x+x+x\right)^{2019}}=\frac{3.x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)
Vậy.......
Từ x3 + y3 + z3 = 3xyz
=> ( x + y + z )( x2 + y2 + z2 - xy - yz - xz ) = 0 ( phân tích như bạn kia )
Vì x + y + z ≠ 0
=> x2 + y2 + z2 - xy - yz - xz = 0
<=> 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz = 0
<=> ( x - y )2 + ( y - z )2 + ( x - z )2 = 0
VT ≥ 0 ∀ x,y,z. Đẳng thức xảy ra <=> x=y=z
Khi đó \(C=\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}=\frac{3x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}\cdot x^{2019}}=\frac{1}{3^{2018}}\)