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

b,

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16 tháng 7 2018

Câu a : Xin phép được sửa đề .

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

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

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

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

Câu b: \(x^2-y^2+10x-6y+16\)

\(=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)

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

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

\(=\left(x-y+2\right)\left(x+y+8\right)\)

Câu c :

\(15-6x-9x^2\)

\(=15-15x+9x-9x^2\)

\(=15\left(1-x\right)+9x\left(1-x\right)\)

\(=\left(1-x\right)\left(15+9x\right)\)

Wish you study well !!

5 tháng 10 2020

a) 5x2 - 5xy + 7y - 7x = ( 5x2 - 5xy ) - ( 7x - 7y ) = 5x( x - y ) - 7( x - y ) = ( x - y )( 5x - 7 )

b) x2 - y2 + 2x + 1 = ( x2 + 2x + 1 ) - y2 = ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )

c) 3x2 + 6xy + 3y2 - 3z2 = 3( x2 + 2xy + y2 - z2 ) = 3[ ( x2 + 2xy + y2 ) - z2 ] = 3[ ( x + y )2 - z2 ] = 3( x + y - z )( x + y + z )

d) ab( x2 + y2 ) + xy( a2 + b2 ) = abx2 + aby2 + a2xy + b2xy

                                                = ( a2xy + abx2 ) + ( aby2 + b2xy )

                                                = ax( ay + bx ) + by( ay + bx )

                                                = ( ay + bx )( ax + by )

17 tháng 10 2017

\(x^2-y^2+10x-6y+16=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)(tách 16 thành 25 - 9 xog nhóm vào)

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

\(=\left(x+y+8\right).\left(x-y+2\right)\)

Chúc bạn học tốt . 

17 tháng 10 2017

  x^2−y^2+10x−6y+16
=(x^2+10x+25)-(y^2+6y+9)

=(x+5)^2-(y+3)^2

=(x+5+y+3)(x+5-y-3)

=(x+y+8)(x-y+2)

24 tháng 7 2016

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

b)\(6x^4+7x^2+2=6x^4+4x^2+3x^2+2=\left(3x^2+2\right)\left(2x^2+1\right)\)

c)\(2x^2-3x-2700=2x^2+72x-75x+2700=\left(2x-75\right)\left(x+36\right)\)

22 tháng 11 2017

a) đề sai thì phải 

b)  2x2 - 2xy - 7x + 7y = (2x2 - 2xy) - (7x - 7y) = 2x(x - y) - 7(x - y) = (x - y)(2x - 7)

c)   x2 - 3x + xy - 3y = (x2 + xy) - (3x + 3y) = x(x + y) - 3(x + y) = (x + y)(x - 3)

đ)    x2 - xy + x - y = (x2 - xy) + (x - y) = x(x - y) + (x - y) = (x - y)(x + 1)

Phối hợp các phương pháp 

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

5 tháng 8 2018

a) \(x^2-6x+8=x^2-4x-2x+8=x\left(x-4\right)-2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)

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

c) \(x^2-x-12=x^2-4x+3x-12=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)

5 tháng 8 2018

a) x^2 -6x +8 

= x^2 - 2x -4x +8

= x(x -2) - 4 ( x-2)

= (x-2)(x-4)

b) x^2 - 4x +3

= x^2 -x - 3x +3

= x(x-1) -3(x-1)

= (x-1)(x-3)

7 tháng 12 2017

c) x^3+3x^2-3x-1=(x^3-1)+(3x^2-3x)=(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+4x+1)

7 tháng 12 2017

\(-25x^6-y^8+10x^3y^4=-\left(5x^3\right)^2-\left(y^4\right)^2+10x^3y^4=-[5x^3-10x^34^4+y^{^{ }4}]=-\left(5y^3-y^4\right)\)

DD
8 tháng 10 2021

a) \(x^2+8x+15=x^2+3x+5x+15=\left(x+3\right)\left(x+5\right)\)

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

c) \(-x^2+7x-6=-x^2+x+6x-6=\left(-x+6\right)\left(x-1\right)\)

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

e) \(2x^2+7x-15=2x^2-3x+10x-15=\left(2x-3\right)\left(x+5\right)\)

a) Ta có: \(x^2+9x+20\)

\(=x^2+4x+5x+20\)

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

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

b) Ta có: \(x^2+x-12\)

\(=x^2+4x-3x-12\)

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

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

c) Ta có: \(6x^2-11x-16\)

\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)

\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)

\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)

\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)

d) Ta có: \(4x^2-8x-5\)

\(=4x^2-10x+2x-5\)

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

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

e) Ta có: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)

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

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

g) Ta có: \(x^3+9x^2+23x+15\)

\(=x^3+x^2+8x^2+8x+15x+15\)

\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+8x+15\right)\)

\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)

\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)

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

h) Ta có: \(2x^4-x^3-9x^2+13x\)

\(=x\left(2x^3-x^2-9x+13\right)\)

i) Ta có: \(x^4+2x^3-16x^2-2x+15\)

\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)

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

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

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

\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)

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