Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x + y = 2
=> ( x + y )2 = 4
<=> x2 + 2xy + y2 = 4
<=> 2xy + 10 = 4
<=> 2xy = -6
<=> xy = -3
Ta có : M = x3 + y3 = ( x + y )( x2 - xy + y2 ) = 2( 10 + 3 ) = 26
Ta có : \(x+y=2\)
\(\Rightarrow\left(x+y\right)^2=4\)
\(\Rightarrow x^2+y^2+2xy=4\)
Mà \(x^2+y^2=10\)
\(\Rightarrow10+2xy=4\)
\(\Rightarrow2xy=-6\)
\(\Rightarrow xy=-3\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10+3\right)=2.13=26\)
Vậy \(x^3+y^3=26\)
(x+y)^3=x^3+y^3+3xy(x+y)=1
=>3xy(x+y)+2=1
=>3xy(x+y)=-1?(vì x+y=1)
=>xy=-1/3=M
b) (x+y)^2=x^2+y^2+2xy=1 =>x^2+y^2=1-2xy=1-2.(-1/3)=5/3
(x^2+y^2)(x^3+y^3)=x^5+y^5 +x^2.y^3+x^3.y^2=x^5+y^5+x^2.y^2(x+y)=...(ráp số vô rồi tính ra kết quả nhé :) )
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Trường hợp x=y=z thì không phải bàn,ns cái trường hợp x+y+z=0
\(\frac{1}{x^2+y^2-z^2}=\frac{1}{\left(x+y\right)^2-2xy-z^2}=\frac{1}{\left(-z\right)^2-z^2-2xy}=\frac{1}{-2xy}\)
Tương tự rồi cộng lại thì \(BT=0\) thì phải
Condition\(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)
Put \(P=\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)
\(=\frac{1}{x^2+\left(y-z\right)\left(y+z\right)}+\frac{1}{y^2+\left(z-x\right)\left(z+x\right)}+\frac{1}{z^2+\left(x-y\right)\left(x+y\right)}\left(4\right)\)
Because \(x^2+y^2+z^2=3xyz\)
\(\Leftrightarrow x^2+y^2+z^2-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)ư\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{cases}}\)
The first case: If \(x+y+z=0\left(1\right)\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}\left(2\right)}\)
From \(\left(1\right)\Rightarrow\hept{\begin{cases}x-y=-2y-z\\y-z=-2z-x\\z-x=-2x-y\end{cases}\left(3\right)}\)
\(\left(2\right)\)and \(\left(3\right)\)into \(\left(4\right)\)we have
\(P=\frac{1}{x^2-x\left(-2z-x\right)}+\frac{1}{y^2-y\left(-2x-y\right)}+\frac{1}{z^2-z\left(-2y-z\right)}\)
\(=\frac{1}{2x^2+2xz}+\frac{1}{2y^2+2xy}+\frac{1}{2z^2+2yz}\)
\(=\frac{1}{2x\left(x+z\right)}+\frac{1}{2y\left(x+y\right)}+\frac{1}{2z\left(z+y\right)}\)
\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)
\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)
\(=\frac{z+x+y}{-2xyz}=0\)( Because x+y+z=0)
The second case:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\left(5\right)\)
We have \(\hept{\begin{cases}\left(x-y\right)^2\ge0;\forall x,y,z\\\left(y-z\right)^2\ge0;\forall x,y,z\\\left(z-x\right)^2\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0;\forall x,y,z\left(6\right)\)
From \(\left(5\right),\left(6\right)\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)
Because \(x=y=z\Rightarrow x^2=y^2=z^2=xy=yz=zx\)
So \(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
\(=\frac{z+x+y}{xyz}=0\)
So...
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 \(\left(x+y\right)^2=4\Rightarrow x^2+y^2+2xy=4\Rightarrow xy=\frac{4-10}{2}=-3\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8-6xy=8-6.\left(-3\right)=26\)
Học tốt!!!!!!
Ta có: x + y = 2
<=> (x + y)2 = 22
<=> x2 + y2 + 2xy = 4
<=> 10 + 2xy = 4
<=> 2xy = -6
<=> xy = -3
Khi đó: M = x3 + y3 = (x + y)(x2 - xy + y2) = 2(10 + 3) = 2.13 = 26