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a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)
b: \(=\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)
d: \(=9x^2+6x+1-9x^2+6x-1=12x\)
a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)
e: \(=x^3+1-x^3+1=2\)
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
\(a,\left(2x-1\right)^2-\left(x-3\right)\left(x+3\right)-1969\\ =4x^2-4x+1-x^2+9-1969\\ =3x^2-4x-1959\)
\(b,\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\\ =4x^2-9y^2-4x^2+4xy-y^2\\ =8y^2+4xy=4y\left(2y+x\right)\)
\(c,\left(x+3y\right)^2+\left(x+y\right)\left(x-y\right)+280\\ =x^2+6xy+9y^2+x^2-y^2+280\\ =2x^2+8y^2+6xy+280\)
a: \(\left(2x-1\right)^2-\left(x-3\right)\cdot\left(x+3\right)-1969\)
\(=4x^2-4x+1-x^2+9-1969\)
\(=3x^2-4x-1959\)
b: \(\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\)
\(=4x^2-9y^2-4x^2+4xy-y^2\)
\(=-10y^2+4xy\)
Lời giải:
a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$
$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.
$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$
$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$
d.
$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$
$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$
$=-x^2y+4x^2-2xy^2-10x$
$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$
\(A=\left(2x+y\right)^2+\left(2x-y\right)^2+\left(4x^2-y^2\right)+3y\\ =\left(4x^2+4xy+y^2\right)+\left(4x^2-4xy+y^2\right)+\left(4x^2-y^2\right)+3y\\ =4x^2+4x^2+4x^2+4xy-4xy+y^2+y^2-y^2+3y=12x^2+3y-y^2\\ B=\left(x-2\right)\left(x+2\right)-\left(x+2\right)^2\\ =\left(x+2\right)\left(x-2-x-2\right)=-4\left(x+2\right)=-4x-8\\ C=\left(3x-4y\right)^2+\left(3x-4y\right)^2\\ =\left(9x^2-24xy+16y^2\right)+\left(9x^2-24xy+16y^2\right)\\ =18x^2-48xy+32y^2\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a) \(\left(x+y\right)^2+\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)
\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2\)
\(=3x^2+y^2\)
b)\(\left(3x+y\right)^2+\left(3x-y\right)^2-\left(2x+y\right)\left(2x-y\right)\)
\(=9x^2+6xy+y^2+9x^2-6xy+y^2-4x^2+y^2\)
\(=14x^2+3y^2\)
c) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2\)
\(=4x^2\)
d)\(-2\left(x^2-9y^2\right)+\left(x-3y\right)^2+\left(x+3y\right)^2\)
\(=\left(x+3y\right)^2-2\left(x+3y\right)\left(x-3y\right)+\left(x-3y\right)^2\)
\(=\left(x+3y-x+3y\right)^2=9y^2\)