Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(\left(x^2+3x+1\right)=a\), ta được:
\(a\left(a+1\right)-6\)\(=a^2+a-6\)\(=\left(a^2+3a\right)-\left(2a+6\right)\)\(=a\left(a+3\right)-2\left(a+3\right)\)
\(=\left(a+3\right)\left(a-2\right)\)
Thay \(a=\left(x^2+3x+1\right)\), ta được:
\(=\left(x^2+3x+1+3\right)\left(x^2+3x+1-2\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
\(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=\left(3x^2-9x\right)-\left(-2x+6\right)\)
\(=3x\cdot\left(x-3\right)-2\cdot\left(x-3\right)\)
\(=\left(3x-2\right)\cdot\left(x-3\right)\)
3x2 - 11x + 6
= 3x2 - 2x - 9x + 6
= ( 3x2 - 2x ) - ( 9x - 6 )
= x( 3x - 2 ) - 3( 3x - 2 )
= ( x - 3 )( 3x - 2 )
\(x.\left(x^2-4\right)-3x+6\)
\(=x.\left(x+2\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x^2+2x\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).[x.\left(x-1\right)+3.\left(x-1\right)]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
\(3x^2+7x-6\)
\(=3x^2+9x-2x-6\)
\(=3x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(3x-2\right)\left(x+3\right)\)
\(\left(x+3\right)^2-\left(2x+6\right)\left(1-3x\right)+\left(3x+1\right)^2\)
\(=x^2+6x+9-\left(2x-6x^2+6-18x\right)+9x^2+6x+1\)
\(=10x^2+12x+10+6x^2+16x-6=16x^2+28x+4\)
\(=4\left(4x^2+7x+1\right)\)
= 3x^2 - 2x-9x+6
=3x^2-9x - (2x-6)
=3x(x-3) - 2(x-3)
= (x-3)(3x-2)
2x3 + 3x2 - 11x - 6
Thử với x = 2 ta có :
2.23 + 3.22 - 11.2 - 6 = 0
Vậy x = 2 là nghiệm của đa thức. Theo hệ quả của định lí Bézout thì đa thức trên chia hết cho x - 2
Thực hiện phép chia 2x3 + 3x2 - 11x - 6 cho x - 2 ta được 2x2 + 7x + 3
=> 2x3 + 3x2 - 11x - 6 = ( x - 2 )( 2x2 + 7x + 3 )
Lại có : 2x2 + 7x + 3 = 2x2 + 6x + x + 3 = 2x( x + 3 ) + ( x + 3 ) = ( x + 3 )( 2x + 1 )
=> 2x3 + 3x2 - 11x - 6 = ( x - 2 )( x + 3 )( 2x + 1 )
\(=2x^4+6x^3-3x^3-9x^2-3x^2-9x+2x+6\)
\(=2x^3\left(x+3\right)-3x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(2x^3-4x^2+x^2-2x-x+2\right)=\left(x+3\right)\left(x-2\right)\left(2x^2+x-1\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(2x^2+2x-x-1\right)=\left(x+3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)\)
2x^4+3x^3-12x^2-7x+6 = (2x^4-x^3)+(4x^3-2x^2)-(10x^2-5x)-(12x-6)
= x^3.(2x-1)+2x^2.(2x-1)-5x.(2x-1)-6.(2x-1) = (2x-1).(x^3+2x^2-5x-6)
= (2x-1).[ (x^3+x^2)+(x^2+x)-(6x+6) ] = (2x-1).(x+1).(x^2+x-6) = (2x-1).(x-1).[(x^2-2x)+(3x-6)]
= (2x-1).(x+1).(x-2).(x+3)
k mk nha
\(3x^2+3x-6\)
\(=3\left(x^2+x-2\right)\)
\(=3\left(x^2-x+2x-2\right)\)
\(=3\left[x\left(x-1\right)+2\left(x-1\right)\right]\)
\(=3\left(x-1\right)\left(x+2\right)\)