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a: \(A=\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)
\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\)
\(=6ab^2\)
b: \(B=\left(2x-1\right)\left(4x^2+2x+1\right)-8\left(x+2\right)\left(x^2-2x+4\right)\)
\(=8x^3-1-8\left(x^3+8\right)\)
\(=8x^3-1-8x^3-64=-65\)
c: \(C=\left(x+2y\right)\left(x^2-2xy+4y^2\right)+\left(2x-y\right)\left(4x^2+2xy+y^2\right)-9x^3\)
\(=x^3+\left(2y\right)^3+\left(2x\right)^3-y^3-9x^3\)
\(=x^3+8y^3+8x^3-y^3-9x^3=7y^3\)
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)
\(=-4.\dfrac{1}{4}+10=9\)
b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)
\(=\left(-2\right).\left(32-32\right)=0\)
a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)
\(=4x^2-4x+1+9-4x^2\)
\(=-4x+10\)
\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)
$=[(x+2)-(x+4)][(x+2)+(x+4)]+x^2-3x+1$
$=(-2)(2x+6)+x^2-3x+1$
$=-4x-12+x^2-3x+1$
$A=x^2-7x-11$
b) $(2x+2)^2-4x(x+2)$$=4(x+1)^2-4x(x+2)$
$=4(x^2+2x+1)-4x^2-8x$
$B=4$
a) \(\dfrac{2x-2y}{x^2-2xy+y^2}=\dfrac{2\left(x-y\right)}{\left(x-y\right)^2}=\dfrac{2}{x-y}\)
b) \(\dfrac{2-2a}{a^3-1}=-\dfrac{2-2a}{1-a^3}=-\dfrac{2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{-2}{\left(1+a+a^2\right)}\)
c) \(\dfrac{x^2-6x+9}{x^2-8x+15}\)
\(=\dfrac{x^2-2.x.3+3^2}{x^2-3x-5x+15}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x^2-3x\right)-\left(5x-15\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)-5\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\)
\(=\dfrac{x-3}{x-5}\)
d) \(\dfrac{x^4-2x^3}{2x^4-x^3}\)
\(=\dfrac{x^3\left(x-2\right)}{x^3\left(2x-1\right)}\)
\(=\dfrac{x-2}{2x-1}\)
\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
\(=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy\right)\)
\(=\left(x+y\right)\left(x+y\right)^2=\left(x+y\right)^3\)
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) $(2x-1)^2-2(2x-3)^2+4$
$=4x^2-4x+1-2(4x^2-12x+9)+4$
$=4x^2-4x+1-8x^2+24x-18+4$
$=-4x^2+20x-13$
b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$
Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:
$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$
$=[(3x+2)-(2y-1)]^2$
$=(3x-2y+3)^2$
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
$=4x^2-4x+1-2(4x^2-12x+9)+4$
$=4x^2-4x+1-8x^2+24x-18+4$
$=-4x^2+20x-13$
b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:
$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$
$=[(3x+2)-(2y-1)]^2$
$=(3x-2y+3)^2$
c) $(x^2+2xy)^2+2(x^2+2xy)y^2+y^4$$=(x^2+2xy)^2+2(x^2+2xy)y^2+(y^2)^2$
$=(x^2+2xy+y^2)^2$
d) $(x-1)^3+3x(x-1)^2+3x^2(x-1)+x^3$
Đặt $a=x-1,\ b=x$:
$=a^3+3a^2b+3ab^2+b^3$
$=(a+b)^3$
$=[(x-1)+x]^3$
$=(2x-1)^3$
e) $(2x+3y)(4x^2-6xy+9y^2)$Dùng $(a+b)(a^2-ab+b^2)=a^3+b^3$:
$=(2x)^3+(3y)^3$
$=8x^3+27y^3$
f) $(x-y)(x^2+xy+y^2)-(x+y)(x^2-xy+y^2)$$=x^3-y^3-(x^3+y^3)$
$=-2y^3$
g) $(x^2-2y)(x^4+2x^2y+4y^2)-x^3(x-y)(x^2+xy+y^2)+8y^3$Dùng $(a-b)(a^2+ab+b^2)=a^3-b^3$:
$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$
$=x^6-8y^3-x^6+x^3y^3+8y^3$
$=x^3y^3$