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Bài làm:
Ta có: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
và
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)
\(=2\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Từ đó thay vào P rút ra:
\(P=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{2020}{2}=1010\)
Vậy P = 1010
\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)
Ta có \(y^3-1=\left(y-1\right)\left(y^2+y+1\right)=-x\left(y^2+y+1\right)\)
(vì \(xy\ne0\Rightarrow x,y\ne0\))
\(\Rightarrow x-1\ne0;y-1\ne0\)
\(\Rightarrow\frac{x}{y^3-1}=\frac{-1}{y^2+y+1}\)
\(x^3-1=\left(x-1\right)\left(x^2-x+1\right)=-y\left(x^2-x+1\right)\Rightarrow\frac{y}{x^3-1}=\frac{-1}{x^2+x+1}\)
\(\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{-1}{y^2+y+1}+\frac{-1}{x^2+x+1}\)
\(=-\left(\frac{x^2+x+1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\right)=-\left(\frac{\left(x+y\right)^2-2xy+\left(x+y\right)+2}{x^2y^2+\left(x+y\right)^2-2xy+xy\left(x+y\right)+xy+\left(x+y\right)+1}\right)\)
\(=-\frac{4-2xy}{x^2y^2+3}\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
\(=\left[\left(\dfrac{-\left(x-y\right)}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}\right):\dfrac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(=\dfrac{-x^2+y^2-x^2-y^2-y+2}{\left(x-2y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)
\(=\dfrac{-2x^2-y+2}{\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)
\(=\dfrac{-1}{x-2y}\)
Thay $x=-1,76$ và $y=\dfrac{3}{25}$ vào $P=\dfrac{-1}{x-2y}$, ta được:
$P=\dfrac{-1}{-1,76-2.(\dfrac{3}{25})}=\dfrac{1}{2}$.
a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)
\(=x^2y\left(4x^3-xy^2+y^3\right)+2xy^2\left(4x^3-xy^2+y^3\right)\)
\(-2y^3\left(4x^3-xy^2+y^3\right)\)
\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)
\(-8x^3y^3+2xy^5-2y^6\)
\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)
\(-\left(x^3y^3+8x^3y^3\right)\)
\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)
b)
(!) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(2x+2y-5\right)\left(x+y-1\right)\)
(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\)
Câu a :
\(VT=\) \(\left(x-1\right)\left(x^2+x+1\right)=x^3-1^3=VP\)
Câu b :
\(VT=\)\(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-y^4=VP\)
Tương tự bạn khai triển là ra nhé
