Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
$6x^2-x-x=6x^2-3x+2x-1=3x(2x-1)+(2x-1)=(2x-1)(3x+1)$
b)
$6x^2-6x-3=3(2x^2-2x-1)$
c)
$15x^2-2x-1=15x^2-5x+3x-1=5x(3x-1)+(3x-1)=(3x-1)(5x+1)$
d)
$2x^3-x^2+5x+3=2x^3+x^2-2x^2-x+6x+3$
$=x^2(2x+1)-x(2x+1)+3(2x+1)=(2x+1)(x^2-x+3)$
e)
$2x^3-5x^2+5x-3=2x^3-3x^2-2x^2+3x+2x-3$
$=x^2(2x-3)-x(2x-3)+(2x-3)=(2x-3)(x^2-x+1)$
a)
$6x^2-x-x=6x^2-3x+2x-1=3x(2x-1)+(2x-1)=(2x-1)(3x+1)$
b)
$6x^2-6x-3=3(2x^2-2x-1)$
c)
$15x^2-2x-1=15x^2-5x+3x-1=5x(3x-1)+(3x-1)=(3x-1)(5x+1)$
d)
$2x^3-x^2+5x+3=2x^3+x^2-2x^2-x+6x+3$
$=x^2(2x+1)-x(2x+1)+3(2x+1)=(2x+1)(x^2-x+3)$
e)
$2x^3-5x^2+5x-3=2x^3-3x^2-2x^2+3x+2x-3$
$=x^2(2x-3)-x(2x-3)+(2x-3)=(2x-3)(x^2-x+1)$
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) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
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) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
\(^{x^3-6x^2-x+30=x^3-5x^2-3x^2+15x-2x^2-10x-6x+30}\)
=x^2(x-5)-3x(x-5)-2x(x-5)-6(x-5)
=(x-5)(x^2-3x-2x-6)
=(x-5)[x(x-3)-2(x-3)]
=(x-5)(x-3)(x-2)
\(x^3-6x^2-x+30\)
= \(x^3-5x^2-3x^2+15x+2x^2-10x-6x+30\)
= \(x^2\left(x-5\right)-3x\left(x-5\right)+2x\left(x-5\right)-6\left(x-5\right)\)
= \(\left(x-5\right)\left(x^2-3x+2x-6\right)\)
= \(\left(x-5\right)\left(x\left(x-3\right)+2\left(x-3\right)\right)\)
= \(\left(x-5\right)\left(x+2\right)\left(x-3\right)\)
a)\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x^2-5x-3x+15\right)\)
\(=\left(x+2\right)\left[x\left(x-5\right)-3\left(x-5\right)\right]\)
\(=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)
nha
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
Mình viết xuôi theo dạng ax2 + bx + c nhé ;-; cho dễ làm
a) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x( 2x + 1 ) + 3( 2x + 1 ) = ( 2x + 1 )( x + 3 )
b) 3x2 - 8x + 4 = 3x2 - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )
c) 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )
d) -6x2 + 7x - 2 = -6x2 + 3x + 4x - 2 = -3x( 2x - 1 ) + 2( 2x - 1 ) = ( 2x - 1 )( 2 - 3x )
e) -3x2 + 7x - 2 = -3x2 + 6x + x - 2 = -3x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 1 - 3x )
f) 2x2 - 5x + 2 = 2x2 - 4x - x + 2 = 2x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 2x - 1 )
g) 3x2 - 8x + 4 = 3x2 - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )
h) 6x2 - 11x + 3 = 6x2 - 2x - 9x + 3 = 2x( 3x - 1 ) - 3( 3x - 1 ) = ( 3x - 1 )( 2x - 3 )
i) 2x2 + 3x - 27 = 2x2 - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 )
j) 4x2 - 5x + 1 = 4x2 - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )
a) \(6x^2-x-1\)
\(=6x^2-3x+2x-1\)
\(=3x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\)
b) \(6x^2-6x-3\)
\(=3\left(2x^2-2x-1\right)\)