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a) \(4x^2-12x+9=\left(2x\right)^2-2\cdot2x\cdot3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2\cdot1\cdot6x+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2\cdot3x\cdot4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
e) \(8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2+2x+1\right)\)
f) \(-8x^3+27=3^3-\left(2x\right)^3=\left(3-2x\right)\left(9+6x+4x^2\right)\)
\(16y^2-4x^2-12x-9=16y^2-\left(2x-3\right)^2\)
\(=\left(4y-2x+3\right)\left(4y+2x-3\right)\)
16y2 - 4x2 - 12x - 9 = 16y2 - (4x2 + 12x + 9) = 16y2 - (2x + 3)2 = (4y - 2x - 3)(4y + 2x + 3)
a) x3 - 4x2 + 12x - 27 = (x - 3)(x2 + 3x + 9) - 4x(x - 3)
= (x - 3)(x2 + 3x + 9 - 4x) = (x - 3)(x2 - x + 9)
b) x3 + 2x2 + 2x + 1 = (x + 1)(x2 - x + 1) + 2x(x + 1)
= (x + 1)(x2 - x + 1 + 2x) = (x + 1)(x2 + x + 1)
c) y4 - 2y3 + 2y - 1 = (y2 - 1)(y2 + 1) - 2y(y2 - 1)
= (y2 - 1)(y2 + 1 - 2y) = (y - 1)(y + 1)(y - 1)2
= (y + 1)(y - 1)3
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
a, \(5y^2-5x^2+6x+6y=5\left(y-x\right)\left(x+y\right)+6\left(x+y\right)\)
\(=\left(x+y\right)\left(5y-5x+6\right)\)
b, \(12x^2+19x+7=12x^2+12x+7x+7\)
\(=12x\left(x+1\right)+7\left(x+1\right)=\left(12x+7\right)\left(x+1\right)\)
\(a,15a^2b^3+5a^3b^2=5a^2b^2\left(3b+a\right)\\ b,x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
a) 15a2b3+5a3b2=5a2b2(3b+a)
b) x2-2x+x-y2=( x2-y2)-(2x+x)
=(x-y)(x+y)-x(2-1)
=(x-y)(x+y)-x3
a) 9x4+16y6-24x2y3
=(3x2)2-2.3x2.4y3+(4y3)2
=(3x2-4y3)2
b) 16x2-24xy+9y2
=(4x)2-2.4x.3y+(3y)2
=(4x-3y)2
c) 36x2-(3x-2)2
=(36x-3x+2)(36x+3x-2)
=(33x+2)(39x-2)
d) 27x3+54x2y+36xy2+8y3
=(3x)3+3.(3x)2.2y+3.3x.(2y)2+(2y)3
=(3x+2y)3
e) y9-9x2y6+27x4y3-27x6
=(y3)3-3.(y3)2.3x2+3.y3.(3x2)2-(3x2)3
=(y3-3x2)3
f) 64x3+1
= (4x)3+13
=(4x+1)[(4x)2-4x.1+12]
=(4x+1)(16x2-4x+1)
e) 27x6-8x3 *sửa đề*
=(3x2)3-(2x)3
=(3x2-2x)[(3x)2+3x2.2x+(2x)2]
=(3x2-2x)(9x2+6x3+4x2)
~~~
a)x2-4y2
=x2-(2y)2
=(x-2y)(x+2y)
b)x3+27y3
=x3+(3y)3
=(x+3y)(x2-3xy+9y2)
c)4x2+12xy+9y2-16
=(2x+3y)2-42
=(2x+3y-4)(2x+3y+4)
d)9x2-24xy+16y2-64
=(3x-4y)2-82
=(3x-4y-8)(3x-4y+8)
e)8x3-27y3
=(2x)3-(3y)3
=(2x-3y)(4x2+6xy+9y2)
f)5x3-7x2+10x-14
=5x3+10x-7x2-14
=5x(x2+2)-7(x2+2)
=(5x-7)(x2+2)