Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
\(x^2=yz\Rightarrow\frac{x}{y}=\frac{z}{x}\left(1\right)\)
\(y^2=xz\Rightarrow\frac{x}{y}=\frac{y}{z}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)
\(\Rightarrow x=y=z\)
Thay y, z bằng x \(\Rightarrow M=\frac{3.x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)
Bổ đề: xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)
Cm:
VT: xyz+(x+y)(y+z)(z+x)
=xyz+xyz+x2z+x2y+y2x+y2z+z2x+z2y+xyz
=xyz+x2z+x2y+xyz+y2z+y2x+xyz+z2x+z2y
=(x+y+z)(xy+yz+xz)
AD bổ đề và đề bài cho
=> (x+y)(y+z)(z+x)=0
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\)
1. x+y=0
ta có x2019+y2019=(x+y)(x2018-x2017y+...+y2018)=0
=> x2019+y2019+z2019=z2019
Có (x+y+z)2019=z2019
=> x2019+y2019+z2019= (x+y+z)2019
Làm tương tự với 2 trường hợp còn lại ta được đpcm
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) 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!
Từ gt suy ra: \(x+\sqrt{x^2+2019}=\dfrac{2019}{y+\sqrt{y^2+2019}}=\sqrt{y^2+2019}-y\).
Tương tự: \(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\).
Do đó dễ dàng suy ra được: \(x+y=0\).
\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).
Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)
\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)
Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)
=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)
= xy+xz+xy+yz+xz+yz
=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)
=4038
Vậy P=4038