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Câu 1: 

a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)

\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)

\(=\left(-3x+14y\right)\left(13x+6y\right)\)

b: \(=\left(0.5x-y\right)^2-\left(2x+2y\right)^2\)

\(=\left(0.5x-y-2x-2y\right)\left(0.5x-y+2x+2y\right)\)

\(=\left(-1.5y-3y\right)\left(2.5x+y\right)\)

c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(0.2x+0.2y\right)^2\)

\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x+2y+\dfrac{1}{5}y+\dfrac{1}{5}x\right)\)

\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x+\dfrac{11}{5}y\right)\)

d: \(=-\left(5x-3\right)^2\)

17 tháng 8 2018

Bài 1:

a) \(25\left(x+2y\right)^2-16\left(2x-y\right)^2\)

\(=\left[5\left(x+2y\right)\right]^2-\left[4\left(2x-y\right)\right]^2\)

\(=\left[5\left(x+2y\right)-4\left(2x-y\right)\right]\left[5\left(x+2y\right)+4\left(2x-y\right)\right]\)

\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)

\(=\left(14y-3x\right)\left(13x+6y\right)\)

b) \(0,25\left(x-2y\right)^2-4\left(x+y\right)^2\)

\(=\left[\dfrac{1}{2}\left(x-2y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)

\(=\left[\dfrac{1}{2}\left(x-2y\right)-2\left(x+y\right)\right]\left[\dfrac{1}{2}\left(x-2y\right)+2\left(x+y\right)\right]\)

\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)

\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)

\(=-3\left(\dfrac{1}{2}x+y\right)\left(\dfrac{5}{2}x+y\right)\)

c) \(\dfrac{4}{9}\left(x-3y\right)^2-0,04\left(x+y\right)^2\)

\(=\left[\dfrac{2}{3}\left(x-3y\right)\right]^2-\left[\dfrac{1}{5}\left(x+y\right)\right]^2\)

\(=\left[\dfrac{2}{3}\left(x-3y\right)-\dfrac{1}{5}\left(x+y\right)\right]\left[\dfrac{2}{3}\left(x-3y\right)+\dfrac{1}{5}\left(x+y\right)\right]\)

\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)

\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)

\(=\dfrac{1}{5}\left(\dfrac{7}{3}x-11y\right).\dfrac{1}{5}\left(\dfrac{13}{3}x-9y\right)\)

\(=\dfrac{1}{25}\left(\dfrac{7}{3}x-11y\right)\left(\dfrac{13}{3}x-9y\right)\)

d) \(-25x^2+30x-9\)

\(=-\left(25x^2-30x+9\right)\)

\(=-\left[\left(5x\right)^2-2.5x.3+3^2\right]\)

\(=-\left(5x-3\right)^2\)

Bài 2:

a) \(x^3y^2-x^2y^3-2x+2y\)

\(=x^2y^2\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2y^2-2\right)\)

Thay x = -1 và y = -2 vào ta được

\(=\left[-1-\left(-2\right)\right]\left[\left(-1\right)^2\left(-2\right)^2-2\right]\)

\(=1\left(4-2\right)\)

\(=2\)

b) \(5x^2-3x+3y-5y^2\)

\(=5\left(x^2-y^2\right)-3\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)

Thay x = 3 và y = 1 vào ta được

\(=5\left(3-1\right)\left(3+1\right)-3\left(3-1\right)\)

\(=5.2.4-3.2\)

\(=34\)

a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)

\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)

\(=\left(-3x+14y\right)\left(13x+6y\right)\)

b: \(=\left(\dfrac{1}{2}x-y\right)^2-\left(2x+2y\right)^2\)

\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)

\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)

c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(\dfrac{1}{5}x+\dfrac{1}{5}y\right)^2\)

\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)

\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)

12 tháng 9 2020

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 )

8 tháng 8 2015

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)

12 tháng 10 2019

2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)

b) \(x^2+16x+64=\left(x+8\right)^2\)

c) \(x^3-8y^3=x^3-\left(2y\right)^3\)

\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)

d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)

12 tháng 7 2019

a,\(xy+3x-7y-21\)

\(=x\left(y+3\right)-7\left(y+3\right)\)

\(=\left(y+3\right)\left(x-7\right)\)

12 tháng 7 2019

\(b,2xy-15-6x+5y\)

\(=\left(2xy-6x\right)+\left(-15+5y\right)\)

\(=2x\left(y-3\right)-5\left(3-y\right)\)

\(=2x\left(y-3\right)+5\left(y-3\right)\)

\(=\left(y-3\right)\left(2x+5\right)\)

4 tháng 8 2017

Mình sửa: Bài 1
2)x2+3x-15

20 tháng 5 2018

a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2

b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)

                         = -(52 – 2 . 5 . x – x2) = -(5 – x)2

c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]

                    = (2x - 1/2)(4x2 + x + 1/4) 

d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)