Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
\(b,6x-6y-x^2+xy\)
\(=\left(6x-6y\right)-\left(x^2-xy\right)\)
\(=6\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left(6-x\right)\)
x3-3x2-3x+1=x3+1-3x2-3x
=(x+1)(x2-x+1)-3x(x+1)
=(x+1)(x2-x+1-3x)
=(x+1)(x2-4x+1)
Câu a :
\(x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=\left(x^2-x\right)-\left(2x-2\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
\(x^3-3x^2+3x-1\) =0
=>\(\left(x-1\right)^3\)=0
=>x-1=0
=>x=1
vậy x =1
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
b ( x^2 + 3x + 2)( x^2 + 7x + 12) - 24
= [ x^2 +x + 2x + 2) ( x^2 +3x + 4x + 12) - 24
= [x(x+1) + 2 (x + 1) [x(x+3) + 4(x+3) ] - 24
= ( x + 1)(x+2) (x+3)(x+4) - 24
= ( x + 1).(x+4) (x+2)(x+3) - 24
=(x^2 + 5x + 4)(x^2+5x+6) - 24
Đặt x^2 + 5x +4 =y ta có:
= y(y+2) - 24
= y^2 + 2y - 24
= y^2 + 2y + 1 - 25
= ( y + 1)^2 - (5)^2
= ( y + 1 - 5 )( y + 1 + 5)
= ( y- 4)(y +6)
Thay y trở lại là đc
đúng nha
\(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)
(3x+1)2-(x+1)2
=[(3x+1)-(x+1)][(3x+1)+(x+1)]
=[3x+1-x-1][3x+1+x+1]
=[2x][4x+2]
=2x*[2(2x+1)]
=4x(2x+1)