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a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
a) 2x.(x2 - 7x - 3)
= 2xx2 + 2x(-7x) + 2x(-3)
= 2x2x - 2.7xx - 2.3x
= 2x3 - 14x2 - 6x
$3x(x+1)-x(3x+2)$
$=3x^2+3x-3x^2-2x$
$=x$
b.$2x(x^2-5x+6)+(x-1)(x+3)$
$=2x^3-10x^2+12x+x^2+2x-3$
$=2x^3-9x^2+14x-3$
a) \(2x\left(x^2-7x-3\right)=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)\cdot4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x^3-4x^2-2x^2+8x+3x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\cdot\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
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 $(a-b)(a^2+ab+b^2)=a^3-b^3$:
$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$
$=x^6-8y^3-x^6+x^3y^3+8y^3$
$=x^3y^3$
a) (2 + xy)2 = 22 + 2.2.xy + (xy)2 = 4 + 4xy + x2y2
b) (5 – 3x)2 = 52 – 2.5.3x + (3x)2 = 25 – 30x + 9x2
c) (5 – x2)(5 + x2) = 52 – (x2)2 = 25 – x4
d) (5x – 1)3 = (5x)3 – 3.(5x)2.1 + 3.5x.12 – 13 = 125x3 – 75x2 + 15x – 1
e) (2x – y)(4x2 + 2xy + y2) = (2x – y)[(2x)2 + 2x.y + y2] = (2x)3 – y3 = 8x3 – y3
f) (x + 3)(x2 – 3x + 9) = (x + 3)(x2 – 3x + 32) = x3 + 33 = x3 + 27
a) (2 + xy)2 = 22 + 2.2.xy + (xy)2 = 4 + 4xy + x2y2
b) (5 – 3x)2 = 52 – 2.5.3x + (3x)2 = 25 – 30x + 9x2
c) (5 – x2)(5 + x2) = 52 – (x2)2 = 25 – x4 d) (5x – 1)3 = (5x)3 – 3.(5x)2.1 + 3.5x.12 – 13 = 125x3 – 75x2 + 15x – 1
e) (2x – y)(4x2 + 2xy + y2) = (2x – y)[(2x)2 + 2x.y + y2] = (2x)3 – y3 = 8x3 – y3
f) (x + 3)(x2 – 3x + 9) = (x + 3)(x2 – 3x + 32) = x3 + 33 = x3 + 27