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27 tháng 9 2020

a, x4 + 2x3 +x2 = x+x+x3 +x2  =(x4+x3 )+(x3 +x) =x3(x +1 ) + x(x+1 ) =(x+1)(x3+x2)

27 tháng 9 2020

a) x4 + 2x3 + x2

= x2(x2 + 2x + 1)

= x2(x + 1)2

= [x(x + 1)]2

= (x2 + x)2

b) 5x3 - 10xy + 5y2 - 20z2

= 5(x3 - 2xy + y2 - 4z2)

c) x2y - xy2 + x3 - y3

= xy(x - y) + (x - y)(x2 + xy + y2)

= (x - y)(x2 + 2xy + y2)

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

d) x2 - xy + 4x - 2y  + 4

= (x2 + 4x + 4) - (xy + 2y)

= (x + 2)2 - y(x + 2)

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

d) x2 - x - 6

= x2 - 3x + 2x - 6

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

= (x + 2)(x - 3)

f) 3x2 - 5x - 8

= 3x2 + 3x - 8x - 8

= 3x(x + 1) - 8(x + 1)

= (3x - 8)(x + 1)

g) x3 + 3x2 + 6x + 4

= (x3 + 3x2 + 3x + 1) + (3x + 3)

= (x + 1)3 + 3(x + 1)

= (x + 1)[(x + 1)2 + 3]

h) 3x3 - 5x2 - 6x + 8

= 3x3 - 3x2 - 2x2 - 6x + 8

= 3x3 - 3x2 - 2x2 + 2x - 8x + 8

= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)

= (3x2 - 2x - 8)(x - 1)

27 tháng 9 2020

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

b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)

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

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

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

c) \(x^2y-xy^2+x^3-y^3\)

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

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

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

27 tháng 9 2020

d) \(x^2-xy+4x-2y+4\)

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

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

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

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

f) \(3x^2-5x-8\)

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

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

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

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

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

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

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

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

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

30 tháng 8 2020

a) x2( x - 1 ) - x + 1

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

= ( x - 1 )( x2 - 1 )

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

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

b) ( a + b )3 - ( a - b )3

= ( a3 + 3a2b + 3ab2 + b3 ) - ( a3 - 3a2b + 3ab2 - b3 )

= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3

= 6a2b + 2b3

= 2b( 3a2 + b )

c) 6x( x - 3 ) + 9 - 3x2

= 6x2 - 18x + 9 - 3x2

= 3x2 - 18x + 9

= 3( x2 - 6x + 3 )

d) x( x - y ) - 5x + 5y

= x( x - y ) - ( 5x - 5y )

= x( x - y ) - 5( x - y )

= ( x - y )( x - 5 )

e) 3( x + 4 ) - x2 - 4x

= 3( x + 4 ) - ( x2 + 4x )

= 3( x + 4 ) - x( x + 4 )

= ( x + 4 )( 3 - x )

f) x2 + 4x - y2 + 4

= ( x2 + 4x + 4 ) - y2

= ( x + 2 )2 - y2

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

g) x2 + 5x

= x( x + 5 )

h) -x2 + 2x + 2y + y2

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

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

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

8 tháng 10 2017

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8 tháng 10 2017

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

\(2,4ax^3-ax=ax\left(4x^2-1\right)=ax\left[\left(2x\right)^2-1^2\right]\) \(=ax\left(2x-1\right)\left(2x+1\right)\)

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

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

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

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

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

\(4,y-4xy+4x^2y\)

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

\(=y\left(1^2-2.1.2x+\left(2x\right)^2\right)^{ }\)

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

17 tháng 6 2017

b1:

câu a,f áp dụng a2-b2=(a-b)(a+b)

câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)

câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)

câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)

câu g xem lại đề

17 tháng 6 2017

b2:

\(f\left(x;y\right)=x^2+y^2-6x+5y+9=\left(x^2-6x+9\right)+\left(y^2+5y+\frac{25}{4}\right)-\frac{25}{4}\)

\(=\left(x-3\right)^2+\left(y+\frac{5}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\)

Dấu "=" xảy ra khi x=3 và y=-5/2

câu c làm tương tự

9 tháng 12 2018

a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)

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

\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)

\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)

\(=\left(x-3\right)\left(2x^2-11x\right)\)

\(=x\left(x-3\right)\left(2x-11\right)\)

b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)

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

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

\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)

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

\(=\left(x-2y+5\right)\left(x+2y+1\right)\)

9 tháng 12 2018

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

\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)

b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)

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