Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0+0+0\)
\(\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
Mà \(\left(x+1\right)^2\ge0\)
\(\left(y+1\right)^2\ge0\)
\(\left(z+1\right)^2\ge0\)
\(\Rightarrow x+1=y+1=z+1=0\)
\(\Rightarrow x=y=z=-1\)
\(\Rightarrow P=1+1+1=3\)
Do x=y=z=-1 nên ;
B=1+1+1=3;
Ban k nha...còn khi nào tìm đc lờ giải mình báo cho bạn..
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)
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 \(P=\left(2.\frac{1}{2}\right)^{2019}=1\)
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy \(VT\ge VP\forall x;y\) để đấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\) thay vào M :
\(M=\left(-1+1\right)^{2015}+\left(1-2\right)^{2016}+\left(-1+1\right)^{2017}=1\)
\(2xy+2x-5z=0\Leftrightarrow z=\frac{2xy+2x}{5}\)
Sau đấy bn thay z vào là ra
Ta có: \(2xy+2x-5z=0\Rightarrow z=\frac{2xy+2x}{5}\)
Thay \(z=\frac{2xy+2x}{5}\)vào A, ta được: \(A=x^2+2y^2+2xy+\frac{8}{5}y+\frac{2xy+2x}{5}+2=x^2+2y^2+\frac{12}{5}xy+\frac{8}{5}y+\frac{2}{5}x+2\)\(=\left(x^2+\frac{12}{5}xy+\frac{36}{25}y^2\right)+\frac{2}{5}\left(x+\frac{6}{5}y\right)+\frac{1}{25}+\left(\frac{14}{25}y^2+\frac{28}{25}y+\frac{14}{25}\right)+\frac{7}{5}\)\(=\left[\left(x+\frac{6}{5}y\right)^2+\frac{2}{5}\left(x+\frac{6}{5}y\right)+\frac{1}{25}\right]+\frac{14}{25}\left(y+1\right)^2+\frac{7}{5}\)\(=\left(x+\frac{6}{5}y+\frac{1}{5}\right)^2+\frac{14}{25}\left(y+1\right)^2+\frac{7}{5}\ge\frac{7}{5}\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}x+\frac{6}{5}y+\frac{1}{5}=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\Rightarrow z=0\)
Ta có: \(x+2y+3x=0\Leftrightarrow x=-\left(2y+3z\right)\)
Lại có: \(2xy+6yz+3xz=0\Leftrightarrow x\left(2y+3z\right)+6yz=0\)
\(\Leftrightarrow-\left(2y+3z\right)\left(2y+3z\right)+6yz=0\Leftrightarrow-\left(2y+3z\right)^2+6yz=0\)
\(\Leftrightarrow\left(2y+3z\right)^2-6yz=0\Leftrightarrow4y^2+12yz+9z^2-6yz=0\)
\(\Leftrightarrow4y^2+6yz+9z^2=0\Leftrightarrow\left(2y+\dfrac{3z}{2}\right)^2+\dfrac{27z^2}{4}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2y+\dfrac{3z}{2}\right)^2=0\\\dfrac{27z^2}{4}=0\end{matrix}\right.\) \(\Rightarrow y=z=0\Rightarrow x=0\)
\(\Rightarrow S=\dfrac{\left(-1\right)^{2019}-1^{2017}+\left(-1\right)^{2015}}{1^{2018}+2.0^{2016}+0^{2014}+2}=\dfrac{-1-1+-1}{1+0+0+2}=\dfrac{-3}{3}=-1\)
Bài 2:
Tìm GTLN: \(x^2+xy+y^2=3\Leftrightarrow xy=\left(x+y\right)^2-3\Rightarrow xy\ge-3\Rightarrow-7xy\le21\)
\(P=2\left(x^2+xy+y^2\right)-7xy\le2.3+21=27\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\xy=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{3},y=-\sqrt{3}\\x=-\sqrt{3},y=\sqrt{3}\end{cases}}\)
Tìm GTNN:
Chứng minh \(xy\le\frac{1}{2}\left(x^2+y^2\right)\Rightarrow\frac{3}{2}xy\le\frac{1}{2}\left(x^2+y^2+xy\right)\)
\(\Rightarrow\frac{3}{2}xy\le\frac{3}{2}\Rightarrow xy\le1\Rightarrow-7xy\ge-7\)
\(P=2\left(x^2+xy+y^2\right)-7xy\ge2.3-7=-1\)
Chúc bạn học tốt.
Làm bài 1 ha :)
Áp dụng BĐT Cô si ta có:
\(\left(1-x^3\right)+\left(1-y^3\right)+\left(1-z^3\right)\ge3\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)
\(\Leftrightarrow\frac{3-\left(x^3+y^3+z^3\right)}{3}\ge\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)
Mặt khác:\(\frac{3-\left(x^3+y^3+z^3\right)}{3}\le\frac{3-3xyz}{3}=1-xyz\)
Khi đó:
\(\left(1-xyz\right)^3\ge\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)\)
Giống Holder ghê vậy ta :D
Ta có : \(x^2+2y+1=0;y^2+2z+1=0;z^2+2x+1=0\)
\(\Rightarrow x^2+2y+1=y^2+2z+1=z^2+2x+1\)
\(\Rightarrow x^2+2y+1-y^2-2z-1-z^2-2x-1=0\)
\(\Rightarrow\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(x-1\right)^2-\left(y-1\right)^2-\left(z+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-1=0\\z+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\\z=-1\end{cases}}\)
Thay \(x=1;y=1;z=-1\)vào A ta có :
\(A=1^{2015}+1^{2016}+\left(-1\right)^{2017}=1+1-1=1\)
Vậy A = 1
Từ \(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)
\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\left(1\right)\)
Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{cases}\left(2\right)}\)
Từ \(\left(1\right)\)và \(\left(2\right)\):
\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+1=0\\y+1=0\\z+1=0\end{cases}}\)
\(\Rightarrow x=y=z=-1\)
\(\Rightarrow A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1+1-1=-1\)
Vậy \(A=-1\)