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Câu 1: Phân tích đa thức thành nhân tử

a) Ta có: \(x^3-x^2y+xy^2\)

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

b) Ta có: \(5a\left(x-y\right)+2b\left(y-x\right)\)

\(=5a\left(x-y\right)-2b\left(x-y\right)\)

\(=\left(x-y\right)\left(5a-2b\right)\)

c) Ta có: \(x\left(x-y\right)-3x+3y\)

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

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

d) Ta có: \(\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)

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

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

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

e) Ta có: \(\left(3x-1\right)^2-16\)

\(=\left(3x-1\right)^2-4^2\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

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

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

f) Ta có: \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4\right)^2-\left(7x\right)^2\)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)

\(=\left(-2x-4\right)\left(12x-4\right)\)

\(=-2\left(x+2\right)\cdot4\cdot\left(3x-1\right)\)

\(=-8\left(x+2\right)\left(3x-1\right)\)

g) Ta có: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+14\right)\left(3x-4\right)\)

h) Ta có: \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)

\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)

\(=\left(4x+7\right)\left(8x+11\right)\)

Câu 3:

Ta có: \(7^{19}+7^{20}+7^{21}\)

\(=7^{19}\left(1+7+49\right)\)

\(=7^{19}\cdot57⋮57\)(đpcm)

23 tháng 10 2016

kết quả thôi nha

23 tháng 10 2016

umk nhanh nha bạn

30 tháng 9 2017

a) x2 - 2x - 4y2 - 4y

= (x2 - 4y2) - (2x + 4y)

= (x + 2y)(x - 2y) - 2(x + 2y)

= (x + 2y)(x - 2y - 2)

= (x + 2y)[x - 2(y + 1)]

b) x4 + 2x3 - 4x - 4

= (x4 - 4) + ( 2x3 - 4x)

= (x2 - 2)(x2 + 2) + 2x(x2 - 2)

= (x2 - 2)(x2 + 2 + 2x)

c) x3 + 2x2y - x -2y

= (x3 - x) + (2x2y - 2y)

= x(x2 - 1) + 2y(x2 - 1)

= (x + 2y)(x2 - 1)

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

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

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

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

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

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

6 tháng 10 2017

b) x8 +7x4+16

= x8+8x4-x4+16

= (x8+8x4+16) - x4

=(x4+4)2-x4

= (x4+4+x2)(x4+4-x2)

c) x5+x-1

= x5 - x4+x3+x4-x3+x2-x2+x-1

= x3(x2-x+1) + x2(x2-x+1) - (x2-x+1)

= (x2-x+1)(x3+x2 -1)

d)x7+x2+1

=x7-x+x2 +x+1

= x (x6-1) + (x2+x+1)

= x(x3-1)(x3+1) + (x2+x+1)

= x(x3+1)(x-1)(x2+x+1)+(x2+x+1)

= (x2+x+1)[x(x3+1)(x-1) +1]

= (x2+x+1)(x5-x4+x2-x+1)

= x (x-1)(x2+x+1)

e) x5+x4+1

= x5+x4+x3 - x3+1

= x3(x2+x+1) - (x-1)(x2+x+1)

= (x2+x+1)(x3-x+1)

f) x8+x+1

= x8-x2+x2+x+1

= x2(x6-1)+(x2+x+1)

= x2(x3-1)(x3+1) +(x2+x+1)

= (x5+x2)(x-1)(x2+x+1) +(x2+x+1)

= (x2+x+1)(x6-x5+x3-x2+1)

23 tháng 11 2017

a) \(x^2-y^2-5x-5y\)

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

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

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

b) \(5x^3-5x^2y-10x^2+10xy\)

\(=\left(5x^3-5x^2y\right)-\left(10x^2-10xy\right)\)

\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x^2-10x\right)\)

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

c) \(x^3-2x^2-x+2\)

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

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

\(=\left(x-2\right)\left(x^2-1\right)\)

\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)

d) \(-y^2+2xy-x^2+3x-3y\)

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

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

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

\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)

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

g) \(4x^2-8x+3\)

\(=4x^2-6x-2x+3\)

\(=\left(4x^2-6x\right)-\left(2x-3\right)\)

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

\(=\left(2x-3\right)\left(2x-1\right)\)

h) \(2x^2-5x-7\)

\(=2x^2+2x-7x-7\)

\(=\left(2x^2+2x\right)-\left(7x+7\right)\)

\(=2x\left(x+1\right)-7\left(x+1\right)\)

\(=\left(x+1\right)\left(2x-7\right)\)

k) \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)

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

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

21 tháng 11 2016

ôi mai dê

21 tháng 11 2016

mấy bài này max dễ bn đăng từng phần 1 mk lm cho

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 )

20 tháng 9 2017

a) 5x-15y=5x-3.5.y=5(x-3y)

c) 14xy(xy+28x)

d) \(\dfrac{2}{7}\left(3x-1\right)\left(x-1\right)\)

e) (x-1)3

f) (x+y-2x)(x+y+2x)=(y-x)(3x+y)

g) (3x+\(\dfrac{1}{2}\))(9x2+\(\dfrac{3}{2}x\)+\(\dfrac{1}{4}\))

h) (x+y-x+y)\(\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

20 tháng 9 2017

2a)

(x+1)(x2+2x)=0

(x+1)x(x+2)=0

\(\left[{}\begin{matrix}x+1=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)