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Lời giải:
a)
$5(2-x)^2+xy-2y=5(x-2)^2+y(x-2)=(x-2)[5(x-2)+y]=(x-2)(5x+y-10)$
b)
$3a^2x-3a^2y+abx-aby=3a^2(x-y)+ab(x-y)$
$=(x-y)(3a^2+ab)=a(x-y)(3a+b)$
c)
$x(x-y)^3-y(y-x)^2-y^2(x-y)=x(x-y)^3-y(x-y)^2-y^2(x-y)$
$=(x-y)[x(x-y)^2-y(x-y)-y^2]$
$=(x-y)(x^3-2x^2y+xy^2-xy)$
$=x(x-y)(x^2-2xy+y^2-y)$
d)
$2ax^3+6ax^2+6ax+18a$
$=2a(x^3+3x^2+3x+9)
$=2a[x^2(x+3)+3(x+3)]$
$=2a(x+3)(x^2+3)$
e) f) Biểu thức không phân tích được thành nhân tử. Bạn xem lại đề.
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
a) Ta có: \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left[4\left(x-2\right)+y\right]\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=\left(x-y\right)\left(3a^2+ab\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
\(=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-yx+y^2-y^2\right]\)
\(=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)\)
d) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=\left(x+3\right)\left(2ax^3+6a\right)\)
\(=2a\left(x+3\right)\left(x^3+3\right)\)
e) Ta có: \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
Bài 2:
a) x(x - 3)- y(3 - x)
= x(x - 3) + y(x - 3)
= (x - 3)(x + y) (1)
Thay x = \(\frac{1}{3}\); y = \(\frac{8}{3}\)vào (1)
Ta có: (\(\frac{1}{3}\)- 3)(\(\frac{1}{3}\)+ \(\frac{8}{3}\))
= \(\frac{-8}{3}\). 3
= -8
a) x(y - x)3 + y(x - y)2 + xy(x - y)
= x(y - x).(y - x)2 + y(x - y)2 + xy(x - y)
= x(y - x)(x - y)2 + y(x - y)2 + xy(x - y)
= (x - y)[x(y - x)(x - y) + y(x - y) + xy]
= (x - y)[x(y - x)(x - y) + y(x - y) + xy]
b) 3a2x - 3a2y + abx - aby
= 3a2(x - y) + ab(x - y)
= a(x - y)(3a + b)
a) x( y - x )3 - y( x - y )2 + xy( x - y )
= -x( x - y )3 - y( x - y )2 + xy( x - y )
= ( x - y )[ -x( x - y )2 - y( x - y ) + xy ]
= ( x - y )[ -x( x2 - 2xy + y2 ) - yx + y2 + xy ]
= ( x - y )( -x3 + 2x2y - xy2 - yx + y2 + xy )
= ( x - y )( -x3 + 2x2y - xy2 + y2 )
b) 3a2x - 3a2y + abx - aby
= 3a2( x - y ) + ab( x - y )
= ( x - y )( 3a2 + ab )
= ( x - y )a( 3a + b )
\(x^3-y^3-36xy\)
\(=\left(x-y\right)^3+3xy\left(x-y\right)-36xy\)
\(=12^3+36xy-36xy\)
\(=1728\)
a. Ta có:
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c+a-b\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
và \(ab^2-ac^2-b^3+bc^2=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
Vậy, \(A=\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{c-a}{-c-b}=\frac{a-c}{c+b}\)
\(a.\: 2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\\ =\left(x+y\right)\left(2a^2b+4a^3b\right)\\ =2a^2b\left(x+y\right)\left(1+2a\right)\)
\(b.\:-3a\left(x-y\right)-a^2\left(7-x\right)\\ =a\left(3y-3x-7a+ax\right)\)
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