a: \(F=-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)

\(=-\left(2x-y\right)^2\cdot\left[2x-y+x\right]-y^3\)

\(=-\left(2x-y\right)^2\cdot\left(3x-y\right)-y^3\)

\(=\left(-4x^2+4xy-y^2\right)\left(3x-y\right)-y^3\)

\(=-12x^3+4x^2y+12x^2y-4xy^2-3xy^2+y^3-y^3\)

\(=-12x^3+16x^2y-7xy^2\)

\(\left(x-2\right)^2+y^2=0\)

mà \(\left(x-2\right)^2+y^2>=0\forall x,y\)

nên dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\)

=>x=2 và y=0

Thay x=2 và y=0 vào F, ta được:

\(F=-12\cdot2^3+16\cdot2^2\cdot0-7\cdot2\cdot0^2\)

\(=-12\cdot2^3\)

\(=-12\cdot8=-96\)

b: \(G=\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=x^3+y^3+3\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)

\(=x^3+y^3+3\left(8x^3-y^3\right)\)

\(=x^3+y^3+24x^3-3y^3\)

\(=25x^3-2y^3\)

Ta có: \(\left\{{}\begin{matrix}x+y=2\\y=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-3\\x=2-y=2-\left(-3\right)=2+3=5\end{matrix}\right.\)

Thay x=5 và y=-3 vào G, ta được:

\(G=25\cdot5^3-2\cdot\left(-3\right)^3\)

\(=25\cdot125-2\cdot\left(-27\right)\)

\(=3125+54=3179\)

c: \(H=\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)

\(=x^3+27y^3+27x^3-y^3\)

\(=28x^3-26y^3\)

Ta có: \(\left\{{}\begin{matrix}3x-y=5\\x=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=3x-5=3\cdot2-5=1\end{matrix}\right.\)

Thay x=2 và y=1 vào H, ta được:

\(H=28\cdot2^3-26\cdot1^3\)

\(=28\cdot8-26\)

=198

\(a,\)\(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2.\)

\(=\left[\left(x-y\right)+\left(x+y\right)\right]^2=\left(x-y+x+y\right)^2=x^2\)

\(b,\)\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(54+8x\right)\)

\(=8x^2-27-54-8x=8x^2-8x-81\)

\(c,\)\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=27x^3+y^3-\left(27x^3-y^3\right)=2y^3\)

\(d,\)\(\left(a+b+c\right)^2-\left(a-c\right)^2-2ab+2bc\)

\(=a^2+b^2+c^2+2ab+2bc+2ac-a^2+2ac-c^2-2ab+2bc\)

\(=b^2+4bc+4ac\)

Yêu cầu của đề là gì ?

Tính

Cái này đơn giản như đang giỡn thôi:

\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)^3-27x^2y\)

\(=\left(3x\right)^3+y^3-\left[\left(3x\right)^3-3.\left(3x^2\right).+3.3x.y^2-y^3\right]-27x^2y\)

\(=27x^3+y^3-27x^3+27x^2y-9xy^2+y^3-27x^2y\)

\(=2y^3-9xy^2\)

\((2x+y) (4x^2-2xy+y^2)-(3x-y)(9x^2+3xy+y^2) =8x^3+y^3-9x^3+y^3=17x^3\)

\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(2x+y\right)\left[\left(2x\right)^2-2xy+y^2\right]-\left(3x-y\right)\left[\left(3x\right)^2+3xy+y^2\right]\)

\(=\left(2x\right)^3+y^3-\left[\left(3x\right)^3-y^3\right]\)

\(=8x^3+y^3-27x^3+y^3\)

\(=-19x^3+2y^3\)

\(=\left(2x\right)^3+y^3+\left(3x\right)^3-y^3-35\left(x^3-1\right)\)

\(=8x^3+27x^3-35x^3+35\)

\(=35x^3-35x^3+35=35\)

\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)-35\left(x-1\right)\left(x^2+x+1\right)\)

\(=8x^3+y^3+27x^3-y^3-35\left(x^3-1\right)\)

\(=35x^3-35x^3+35\)

\(=35\)