Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
\(x^2+y^2-2x-4y+5=0\)
\(\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)=0\)
\(\Leftrightarrow (x-1)^2+(y-2)^2=0\)
Vì $(x-1)^2; (y-2)^2\geq 0$ với mọi $x,y\in\mathbb{R}$ nên để tổng của chúng bằng $0$ thì $(x-1)^2=(y-2)^2=0$
$\Rightarrow x=1; y=2$
Vậy...........
Bài 2:
Ta có:
\(a(a-b)+b(b-c)+c(c-a)=0\)
\(\Leftrightarrow 2a(a-b)+2b(b-c)+2c(c-a)=0\)
\(\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\)
Lập luận tương tự bài 1, ta suy ra :
\((a-b)^2=(b-c)^2=(c-a)^2=0\Rightarrow a=b=c\)
Khi đó, thay $b=c=a$ ta có:
\(P=a^3+b^3+c^3-3abc+3ab-3c+5\)
\(=3a^3-3a^3+3a^2-3a+5=3a^2-3a+5\)
\(=3(a^2-a+\frac{1}{4})+\frac{17}{4}=3(a-\frac{1}{2})^2+\frac{17}{4}\geq \frac{17}{4}\)
Vậy $P_{\min}=\frac{17}{4}$
Giá trị này đạt được tại $b=c=a=\frac{1}{2}$
c)\(x^3+3xy+y^3\)
\(=x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=\left(x^2-xy+y^2\right)+3xy\)
\(=x^2-xy+y^2+3xy\)
\(=x^2+2xy+y^2=\left(x+y\right)^2\)
\(=1^2=1\)
\(x+y+z=6\)
\(\Rightarrow\left(x+y+z\right)^2=36\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=36\)
\(\Rightarrow2xy+2yz+2zx=24\)
\(\Rightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y-z\right)^2\ge0\forall y,z\\\left(z-x\right)^2\ge0\forall z,x\end{matrix}\right.\)
Nên \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-y\right)^2=\left(y-z\right)^2=\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow x=y=z=2\)
\(M=\left(x+y+z\right)^2+\left(y+z\right)^2-2\left(y+z\right)\left(x+y+z\right)=\left[\left(x+y+z\right)-\left(y+z\right)\right]^2=x^2\)\(N=\left(x-1\right)^3+\left(x+1\right)^3=\left[\left(x-1\right)+\left(x+1\right)\right]\left[\left(x-1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)=\(2x\left(x^2-2x+1-x^2+1+x^2+2x+1\right)=2x\left(2x+3\right)\)
a, \(M=\left(x+y+z\right)^2+\left(y+z\right)^2-2\left(y+z\right)\left(x+y+z\right)\)
\(=x^2+y^2+z^2+2xy+2yz+2xz+y^2+2yz+z^2-2\left(xy+y^2+yz+xz+yz+z^2\right)\)
\(=x^2+2y^2+2z^2+2xy+4yz+2xz-2xy-2y^2-2yz-2xz-2yz-2z^2\)
\(=x^2\)
b, \(N=\left(x-1\right)^3+\left(x+1\right)^3\)
\(=x^3-3x^2+3x-1+x^3+3x^2+3x+1\)
\(=2x^3+6x\)
mình sửa đề nhé~
Có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x;y;z\)
\(\Rightarrow2.\left(x^2+y^2+z^2\right)-2xy-2yz-2xz\ge0\forall x;y;z\)
\(\Leftrightarrow2.\left(x^2+y^2+z^2\right)\ge2xy+2yz+2xz\forall x;y;z\)
\(\Leftrightarrow3.\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2xy+2yz+2xz\forall x;y;z\)
\(\Leftrightarrow3.\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\forall x;y;z\)
Mà \(3.\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)
\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\x=z\end{matrix}\right.\Leftrightarrow x=y=z\)
Có: \(x^{2018}+y^{2018}+z^{2018}=27^{673}\)
\(\Leftrightarrow3.x^{2018}=27^{673}\)
\(\Leftrightarrow x^{2018}=3^{2018}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
đến đây bạn tự làm nốt nhé
\(x^2y-y^2z+x^2z-z^2x+y^2z+z^2y=2xyz\\ \Leftrightarrow\left(x-y\right)\left(y+z\right)\left(z-x\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+z=0\\z-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-z\\z=x\end{matrix}\right.\left(dpcm\right)\)
Có: x2+y2+z2≥1/3 (x+y+z)2 =4/3
=> x2+y2+z2 -3 >= 4/3 - 3 = -5/3
Dấu "=" xảy ra khi x=y=z=2/3