Cho 3x+3y=2021. Tính gt bt sau một cách hợp lí: P=x3+2021xy+y3
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\(P=x^3+y^3+2021xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+2021xy\)
\(=\left(\dfrac{2021}{3}\right)^3-2021xy+2021xy\)
\(=\dfrac{8254655261}{27}\)
\(3x+3y=2021\)
\(\Leftrightarrow x+y=\dfrac{2021}{3}\)
\(P=x^3+y^3+2021xy\)
\(=\left(x+y\right)^3-3xy\cdot\left(x+y\right)+2021xy\)
\(=\left(\dfrac{2021}{3}\right)^3-3xy\cdot\dfrac{2021}{3}+2021xy\)
\(=\dfrac{8254655261}{27}\)
\(A=2x^2+y^2-2x+2xy+2y+3=y^2+2y\left(x+1\right)+\left(x+1\right)^2+\left(x^2-4x+4\right)-2=\left(y+x+1\right)^2+\left(x-2\right)^2-2\ge-2\)
\(minA=-2\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
\(P=x^3+2021xy+y^3\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+2021xy\)
\(=\left(\dfrac{2021}{3}\right)^3\)
\(=\dfrac{8254655261}{27}\)
Ta có: \(x^3-y^3=3x-3y\Leftrightarrow x^2+xy+y^2=3\) (Do \(x\neq y\)).
Tương tự: \(y^2+yz+z^2=3;z^2+zx+x^2=3\).
Cộng vế với vế ta có: \(2\left(x^2+y^2+z^2\right)+xy+yz+zx=9\)
\(\Leftrightarrow\dfrac{3\left(x^2+y^2+z^2\right)}{2}+\dfrac{\left(x+y+z\right)^2}{2}=9\).
Mặt khác, từ đó ta cũng có: \(\left(x^2+xy+y^2\right)-\left(y^2+yz+z^2\right)=0\Leftrightarrow\left(x+y+z\right)\left(x-z\right)=0\Leftrightarrow x+y+z=0\).
Do đó \(x^2+y^2+z^2=6\left(đpcm\right)\).
a) \(x^3-3x+3y-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)=\left(x-y\right)\left(x^2+xy+y^2-3\right)\)
\(a,=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\\ =\left(x-y\right)\left(x^2+xy+y^2-3\right)\\ b,=\left(x+2\right)^2-\left(x-1\right)^2\\ =\left(x+2-x+1\right)\left(x+2+x-1\right)\\ =3\left(2x+1\right)\)
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
c) \(3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)d) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
\(c,=3\left(x+y\right)-\left(x+y\right)^2=\left(3-x-y\right)\left(x+y\right)\\ d,=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(P=x^3+y^3+2021xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+2021xy\)
\(=\left(\dfrac{2021}{3}\right)^3=\dfrac{8254655261}{27}\)