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6x^2 + 15x - 36
= 6x^2 + 24x - 9x - 36
= (6x^2 + 24x) - (9x + 36)
= 6x(x + 4) - 9(x+4)
= (6x - 9) (X + 4)
= 3(2x - 3)(x + 4)
\(6x^2+15-36\)
\(=3\left(2x^2+5x-12\right)\)
\(=3\left(2x^2+8x-3x-12\right)\)
\(=3\left[2x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=3\left(2x-3\right)\left(x+4\right)\)
a)3x2+7x-6
=3x2-2x+9x-6
=x(3x-2)+3(3x-2)
=(x+3)(3x-2)
b)8x2-2x-3
=8x2-6x+4x-3
=2x(4x-3)+(4x-3)
=(2x+1)(4x-3)
c)6x2-15x+6
=3(2x2-5x+2)
=3(2x2-x-4x+2)
=3[x(2x-1)-2(2x-1)]
=3(x-2)(2x-1)
d)10x2+7x-6
=10x2+12x-5x-6
=2x(5x+6)-(5x+6)
=(2x-1)(5x+6)
a) Ta có \(6x^2-15x+9=3.\left(2x^2-5x+3\right)=3.\left(2x^2-2x^2-3x+3\right)\)
\(=3.\left[2x.\left(x-1\right)-3.\left(x-1\right)\right]=3.\left(2x-3\right).\left(x-1\right)\)
b) Ta có \(5x^2+12x+7=5x^2+5x+7x+7=5x.\left(x+1\right)+7.\left(x+1\right)\)
\(=\left(x+1\right)\left(5x+7\right)\)
a) \(6x^2-15x+9\)
\(=3\left(2x^2-5x+3\right)\)
\(=3\left(x-1\right)\left(2x-3\right)\)
b) \(5x^2+12x+7\)
\(=\left(5x^2+5x\right)+\left(7x+7\right)\)
\(=5x\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(5x+7\right)\)
a. = \(6x^2+3x+12x+6=3x\left(2x+1\right)+6\left(2x+1\right)=\left(2x+1\right)\left(3x+6\right)=3\left(2x+1\right)\left(x+2\right)\)
b. \(=6x^2-3x-12x+6=3x\left(2x-1\right)-6\left(2x-1\right)=3\left(2x-1\right)\left(x-2\right)\)
c. \(=6x^2+2x+18x+6=2x\left(3x+1\right)+6\left(3x+1\right)=\left(2x+6\right)\left(3x+1\right)=2\left(x+3\right)\left(3x+1\right)\)
d. \(=6x^2-2x-18x+6=2x\left(3x-1\right)-6\left(3x-1\right)=2\left(x-3\right)\left(3x-1\right)\)
(+) 6x^2 + 15x + 6 = 6x^2 + 12x + 3x + 6
= 6x ( x+ 2 ) + 3 ( x + 2)
= ( 6x + 3) ( x + 2 )
= 3 (2 x + 1 ) ( x + 2)
(+) 6x^2 - 15x + 6 = 6x^2 - 3x - 12x + 6
= 3x( 2x - 1 ) - 6 ( 2x - 1 )
= ( 3 x - 6 )( 2x - 1)
= 3 ( x- 2 ) (2x- 1)
6x^5+15x^4+20x^3+15x^2+6x+1
=3x^4(2x+1)+6x^3(2x+1)+7x^2(2x+1)+4x(2x+1)+(2x+1)
=(2x+1)(3x^4+6x^3+7x^2+4x+1)
=(2x+1)(3x^2(x^2+x+1)+3x(x^2+x+1)+(x^2+x+1)
=(2x+1)(x^2+x+1)(3x^2+3x+1)
Bài làm ai trên 11 điểm tích mình thì mình tích lại
Ông tùng hơn tùng số tuổi là :
29 + 32 = 61 (tuổi )
Vậy ông của tùng hơn tùng 61 tuổi
b) \(B=\left(2x^2+x-2\right)\left(2x^2+x-3\right)-12\)
Đặt \(2x^2+x-2=t\)
Ta được:
\(B=t\left(t-1\right)-12\)
\(B=t^2-t-12\)
\(B=t^2+3t-4t-12\)
\(B=t\left(t+3\right)-4\left(t+3\right)\)
\(B=\left(t+3\right)\left(t-4\right)\)
\(B=\left(2x^2+x+1\right)\left(2x^2+x-6\right)\)
\(B=\left(2x^2+x+1\right)\left(2x^2+4x-3x-6\right)\)
\(B=\left(2x^2+x+1\right)\left[2x\left(x+2\right)-3\left(x+2\right)\right]\)
\(B=\left(2x^2+x+1\right)\left(x+2\right)\left(2x-3\right)\)
Vậy...
=6x^2 - 4x - 9x +6
=(6x^2 -4x) - (9x-6)
=2x(3x -2) - 3(3x-2)
=(3x-2) (2x - 3)
6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)
7) \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)
8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)
9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)
10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)
6) 9x3y2 + 3x2y2 = 3x2y2( 3x + 1 )
7) x3 + 2x2 + 3x = x( x2 + 2x + 3 )
8) 6x2y + 4xy2 + 2xy = 2xy( 3x + 2y + 1 )
9) 5x2( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )
10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )
=6x^2-12x-3x+6
=6x(x-2)-3(x-2)
=(6x-3)(x-2)
ừ hè quên mất 12 với 3