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a) 2x2-6x-x+3 = 2x(x-3) - (x-3) = (x-3)(2x-1)
b) x2-x-5x+5 = x(x-1) - 5(x-1) = (x-1)(x-5)
c) 5x(x-2y) + 2( x-2y)2 = (x-2y)(5x+2x-2y) = (x-2y)(7x-2y)
chú ý : (A-B)2=(B-A)2
d) 7x(4-y)2 - (4-y)3 = ( 16-8y+y2) (7x-4+y)
a) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(2x-1\right)\)
b) \(x^2-6x+5=x^2-5x-x+5=x\left(x-5\right)-\left(x-5\right)=\left(x-5\right)\left(x-1\right)\)
c)\(5x\left(x-2y\right)+2\left(2y-x\right)^2=5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
d) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)+\left(y-4\right)^3=\left(y-4\right)\left(7x-y-4\right)\)
Câu 1:
$(2x^2-3)(x+5)=2x^2(x+5)-3(x+5)=2x^3+10x^2-3x-15$
Câu 2:
a.
$(x+3)^2=x^2+2.x.3+3^2=x^2+6x+9$
b.
$y^2-25=y^2-25$
Áp dụng HĐT a2 - b2 = ( a - b )( a + b )
và tính chất an.bn = ( a.b )n ( với n ∈ N* )
a) ( 3x + 1 )2 - ( x + 1 )2
= [ ( 3x + 1 ) - ( x + 1 ) ][ ( 3x + 1 ) + ( x + 1 ) ]
= ( 3x + 1 - x - 1 )( 3x + 1 + x + 1 )
= 2x( 4x + 2 )
= 2x.2( 2x + 1 )
= 4x( 2x + 1 )
b) ( x + y )2 - ( x - y )2
= [ ( x + y ) - ( x - y ) ][ ( x + y ) + ( x - y ) ]
= ( x + y - x + y )( x + y + x - y )
= 2y.2x = 4xy
c) ( 2xy + 1 )2 - ( 2x + y )2
= [ ( 2xy + 1 ) - ( 2x + y ) ][ ( 2xy + 1 ) + ( 2x + y ) ]
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )9 y + 1 )( 2x + 1 )
d) 9( x - y )2 - 4( x + y )2
= 32( x - y )2 - 22( x + y )2
= [ 3( x - y ) ]2 - [ 2( x + y ) ]2
= ( 3x - 3y )2 - ( 2x + 2y )2
= [ ( 3x - 3y ) - ( 2x + 2y ) ][ ( 3x - 3y ) + ( 2x + 2y ) ]
= ( 3x - 3y - 2x - 2y )( 3x - 3y + 2x + 2y )
= ( x - 5y )( 5x - y )
e) ( 3x - 2y )2 - ( 2x - 3y )2
= [ ( 3x - 2y ) - ( 2x - 3y ) ][ ( 3x - 2y ) + ( 2x - 3y ) ]
= ( 3x - 2y - 2x + 3y )( 3x - 2y + 2x - 3y )
= ( x + y )( 5x - 5y )
= ( x + y )5( x - y )
f) ( 4x2 - 4x + 1 ) - ( x + 1 )2
= ( 2x - 1 )2 - ( x + 1 )2
= [ ( 2x - 1 ) - ( x + 1 ) ][ ( 2x - 1 ) + ( x + 1 ) ]
= ( 2x - 1 - x - 1 )( 2x - 1 + x + 1 )
= 3x( x - 2 )
x^2 + 3xy+2y^2 = x^2 +2xy+y^2+xy+y^2=(x+y)^2 + y(x+y)=(x+y)(x+2y)
a)x4-1=(x2-1)(x2+1)=(x-1)(x+1)(x2+1)
b)x2-y2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)
c)x2-6x-y2+9=(x2-6x+9)-y2=(x-3)2-y2=(x-y-3)(x+y-3)
d)5x2+3(x+y)2-5y2
=5(x2-y2)+3(x+y)2
=5(x-y)(x+y)+3(x+y)2
=(x+y)(5x-5y+3x+3y)
=(x+y)(8x-2y)
a) \(x^2-2x-y^2+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
b) \(2x+2y-x^2-xy\)
\(=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)
x2-y2 -2x+2y=(x+y)(x-y)-2(x-y)=(x-y)(x-y-2)
2(x+y)-x(x+y)=(x+y)(2-x)
a)\(\left(x-y\right)^2-2\left(x-y\right)+1=\left(x-y-1\right)^2\)
b)\(x^2-2y-1-2x+1-y^2=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left[\left(x-1\right)-\left(y+1\right)\right]\left[\left(x-1\right)+\left(y+1\right)\right]\)
\(=\left(x-y-2\right)\left(x+y\right)\)
c)\(x^2-y^2-2x-1=x^2-\left(y^2+2x+1\right)\)
\(=x^2-\left(y+1\right)^2\)
\(=\left(x^2-y-1\right)\left(x^2+y+1\right)\)
A. Ta có: (x - y)2 - 2(x - y)+1 = (x - y)2 - 2.(x - y).1 +12 = ( x - y - 1)2
B. Ta có: x2 - 2y -1 - 2x +1 -y2 = (x2 - y2) - (2x - 2y) -1+1 = (x - y)(x + y) - 2(x - y) = (x - y)(x + y - 2)
C. Ta có: x2 - y2 -2y -1 = x2 -(y2 - 2y -1) = x2 - ( y2 +2y1 + 1) = x2 - (y+1)2 = (x - y - 1)(x + y +1)
k cho mình nha bạn hihj!!! ~3~
a)Bt = (x2-a2)-(2x-2a)
=....
b)Bấm máy tìm nghiệm đi rồi phân tích
c);d);e);f)Nhóm số đầu vs số thứ 2, số thứ 3 vs số thứ 4
ta có :