\(12x^3+4x^2-27x-9\)

b)

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AH
Akai Haruma
Giáo viên
9 tháng 8 2018

c)

\(30ax-34bx-15a+17b\)

\(=(30ax-15a)-(34bx-17b)\)

\(=15a(2x-1)-17b(2x-1)\)

\(=(2x-1)(15a-17b)\)

d)

\(x^3-x^2y-x^2z-xyz\)

\(=x[x^2-xy-xz-yz]\)

AH
Akai Haruma
Giáo viên
9 tháng 8 2018

a)

\(12x^3+4x^2-27x-9\)

\(=(12x^3+4x^2)-(27x+9)\)

\(=4x^2(3x+1)-9(3x+1)\)

\(=(3x+1)(4x^2-9)=(3x+1)[(2x)^2-3^2]\)

\(=(3x+1)(2x-3)(2x+3)\)

b)

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

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

\(=x^2[x^2(x^2-1)+2(x+1)]\)

\(=x^2[x^2(x-1)(x+1)+2(x+1)]\)

\(=x^2(x+1)[x^2(x-1)+2]\)

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

\(=x^2(x+1)[(x+1)(x^2-x+1)-(x-1)(x+1)]\)

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

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

29 tháng 6 2017

a) \(12x^5y+24x^4y^2+12x^3y^3\)

\(=12x^3y\left(x^2+2xy+y^2\right)\)

\(=12x^3y\left(x+y\right)^2\)

b) \(x^2-2xy-4+y^2\)

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

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

g) \(12xy-12xz+3x^2y-3x^2z\)

\(=12x\left(y-z\right)+3x^2\left(y-z\right)\)

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

e) \(16x^2-9\left(x^2+2xy+y^2\right)\)

\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)

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

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

d) làm tương tự như phần g chỉ khác là phải nhóm( nhóm xen kẽ), phần f cũng vậy

Câu 2 nha

\(a,x^4+2x^3+x^2\)

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

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

\(c,x^2-x+3x^2y+3xy^2+y^3-y\)

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

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

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

19 tháng 1 2020

a) \(12x-9-4x^2\)

\(=-\left(4x^2-12x+9\right)\)

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

b)\(1-9x+27x^2-27x^3\)

\(=\left(1-3x\right)^{^3}\)

c)\(\frac{x^2}{4}+2xy+4y^2-25\)

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

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

d)\(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15\)

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

\(=\left(x^2-4x\right)\left(x^2-4x-3\right)-5\left(x^2-4x-3\right)\)

\(=\left(x^2-4x-5\right)\left(x^2-4x-3\right)\)

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

\(=\left[x\left(x+1\right)-5\left(x+1\right)\right]\left(x^2-4x-3\right)\)

\(=\left(x-5\right)\left(x+1\right)\left(x^2-4x-3\right)\)

Chúc bạn học tốt !

20 tháng 4 2017

a) x2 – 4 + (x – 2)2

= (x2 – 22) + (x – 2)2 = (x – 2)(x + 2) + (x – 2)2

= (x – 2) [(x + 2) + (x – 2)]

= (x – 2)(x + 2 + x – 2)

= 2x(x – 2)

b) x3 – 2x2 + x – xy2

= x(x2 – 2x + 1 – y2) = x[(x2 – 2x + 1) – y2]

= x[(x – 1)2 – y2]

= x[(x – 1) + y] [(x – 1) – y]

= x(x – 1 + y)(x – 1 – y)

c) x3 – 4x2 – 12x + 27

= (x3 + 27) – 4x(x + 3)

= (x + 3)(x2 – 3x + 9) – 4x(x + 3)

= (x + 3)(x2 – 3x + 9 – 4x)

= (x + 3)(x2 – 7x + 9)

8 tháng 8 2018

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

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

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

\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)

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

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

\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)

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

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

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

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

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

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

8 tháng 8 2018

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

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

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

\(x^3-4x^2+12x-27\)

\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)

\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)

\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)

\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)

\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)

18 tháng 9 2020

27x6 + 125y6 = ( 3x2 )3 + ( 5y2 )3 = ( 3x2 + 5y2 )( 9x4 - 15x2y2 + 25y4 )

8a6 - 8b6 = ( 2a2 )3 - ( 2b2 )3 = ( 2a - 2b )( 4a2 + 4ab + 4b2 ) = 2( a - b )4( a2 + ab + b2 ) = 8( a - b )( a2 + ab + b2 )

x4 + 64y4 = x4 + 16x2y2 + 64y4 - 16x2y2 

                = ( x4 + 16x2y2 + 64y4 ) - 16x2y2

                = ( x2 + 8y2 )2 - ( 4xy )2

                = ( x2 + 8y2 - 4xy )( x2 + 8y2 + 4xy )

x4 + x3 + 2x2 + x + 1 = x4 + x3 + x2 + x2 + x + 1

                                  = ( x4 + x3 + x2 ) + ( x2 + x + 1 )

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

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

\(27x^6+125y^6=\left(3x^2\right)^3+\left(5y^2\right)^3=\left(3x^2+5y^2\right)\left(9x^4-15x^2.y^2+25y^4\right)\)

\(8a^6-8b^6=8\left(a^6-b^6\right)=8\left(\left(a^3\right)^2-\left(b^3\right)^2\right)=8\left(a^3-b^3\right)\left(a^3+b^3\right)\)

                                                       \(=8\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(x^{\text{4}}+64y^4=x^4+64y^4+16x^2y^2-16x^2y^2\)

                       \(=\left(8y^2+x^2\right)^2-\left(4xy\right)^2=\left(8y^2+x^2+4xy\right)\left(8y^2+x^2-4xy\right)\)

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

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