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b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)
c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)
Bài 1
a) (15x4 - 8x3 + 3x2) : 3x2 = 15x4 : 3x2 - 8x3 : 3x2 + 3x2 : 3x2 = 5x2 - \(\dfrac{8}{3}x\) + 1
b) (4x2 - 4xy + y2) : (2x - y) = (2x - y)2 : (2x - y) = 2x - y
c) dùng phương pháp chia đa thức 1 biến đã sắp xếp nha
d) (x2 - 2x + 1) : (x - 1) = (x - 1)2 : (x - 1) = x - 1
Bài 2
a) x2 + 3x + 3xy + 9xy = x2 + 3x + 12xy = x (x + 3 + 4y)
b) x2 - y2 + 2x + 1 = x2 + 2x + 1 - y2 = (x + 1)2 - y2 = (x + 1 - y)(x + 1 + y)
c) x2 - xy + x - y = x (x - y) + (x - y) = (x - y)(x + 1)
x4,x2,y2 là mũ nhá các bn
giúp mk nhanh lên mình sắp phải nộp r
\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)
Lời giải:
a) (x2 + 2xy + y2) : (x + y)
= (x + y)2 : (x + y)
= x + y
b) (125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1
c) (x2 – 2xy + y2) : (y – x)
= (x – y)2 : [-(x – y)]
= -(x – y)
= y – x
Hoặc (x2 – 2xy + y2) : (y – x)
= (y2 – 2yx + x2) : (y – x)
= (y – x)2 : (y – x)
= y – x
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 $1-2y=-(2y-1)$:
$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$
$=[(3x+2)-(2y-1)]^2$
$=(3x-2y+3)^2$
1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x2-y2+x+y+x-y)=2(x2-y2+2x)=2x2-2y2+4x
2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)
b.=(3x)2+2.3x.y+y2-(2z)2=(3x+y)2-(2z)2=(3x+y-2z)(3x+y+2z)
c.=x2-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)
\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
hk tốt
^^
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 $(a-b)(a^2+ab+b^2)=a^3-b^3$:
$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$
$=x^6-8y^3-x^6+x^3y^3+8y^3$
$=x^3y^3$
(x2 – 2xy + y2) : (y – x)
= (x – y)2 : [-(x – y)]
= -(x – y)
= y – x
Hoặc (x2 – 2xy + y2) : (y – x)
= (y2 – 2yx + x2) : (y – x)
= (y – x)2 : (y – x)
= y – x
a ) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)\)
\(=\left(2x+y\right)^2:\left(2x+y\right)\)
\(=2x+y\)
b ) \(\left(27x^3+1\right):\left(3x+1\right)\)
\(=\left(3x+1\right)\left(9x^2-3x+1\right):\left(3x+1\right)\)
\(=9x^2-3x+1\)
c ) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)\)
\(=\left(x-3y\right)^2:\left(3y-x\right)\)
\(=\left(3y-x\right)^2:\left(3y-x\right)\)
\(=3y-x\)
d ) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)\)
\(=2x-1\)
:D