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\(ab^3c^2-a^2b^2c^2-ab^2c^3+a^2bc^3=abc^2\left(b^2-ab-bc+ac\right)=abc^2\left(b-a\right)\left(b-c\right)\)
Ta có
a b 3 c 2 − a 2 b 2 c 2 + a b 2 c 3 − a 2 b c 3 = a b c 2 ( b 2 – a b + b c – a c ) = a b c 2 [ ( b 2 – a b ) + ( b c – a c ) ] = a b c 2 [ b ( b – a ) + c ( b – a ) ] = a b c 2 ( b + c ) ( b – a )
Vậy ta cần điền b – a
Đáp án cần chọn là: A
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
Bài làm
-x2 - y2 + xy + 16
= -( x2 - xy + y2 ) + 16
= -( x - y )2 + 42
= -[ ( x - y )2 - 42 ]
= - [ ( x - y - 4 )( x - y + 4 ) ]
# Học tốt #
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
=xy ( x + y ) + z ( x^2 + 2xy + y^2 ) = xy ( x + y ) + z ( x + y ) ^ 2 = ( x + y ) ( xy + xz + yz )
\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
Cách 1: x 2 + 2xy - 15 y 2 = ( x 2 + 2xy + y 2 ) - 16 y 2
= x + y 2 - 4 y 2
= (x + y + 4y)(x + y – 4y)
= (x + 5y)(x – 3y).
Cách 2: x 2 + 2xy - 15 y 2 = x 2 + 5xy – 3xy - 15 y 2
= x(x + 5y) – 3y(x + 5y)
= (x – 3y)(x + 5y).
a,\(5xy^3-2xyz-15y^2+6z\)
\(=\left(5xy^3-15y^2+6z-2xyz\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)\)
\(=\left(5y^2-2z\right)\left(xy-3\right)\)
a) 5xy3 - 2xyz - 15y2 + 6z
= ( 5xy3 - 15y2 ) - ( 2xyz - 6z )
= 5y2( xy - 3 ) - 2z( xy - 3 )
= ( xy - 3 )( 5y2 - 2z )
b) ab3c2 - a2b2c2 + ab2c3 - a2bc3
= abc2( b2 - ab + bc - ac )
= abc2[ ( b2 - ab ) + ( bc - ac ) ]
= abc2[ b( b - a ) + c( b - a ) ]
= abc2( b - a )( b + c )