Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(x^2 + x)^2 - 2.(x^2 + x) - 15
= (x2 + x)2 - 2(x2 + x) + 1 - 16
= (x2 + x + 1)2 - 16
= (x2 + x + 1 - 4)(x2 + x + 1 + 4)
= (x2 + x - 3)(x2 + x + 5)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(\left(x^2+x\right)^2-2\left(x^2+x\right)+1\right)-16\)
\(=\left(x^2+x-1\right)^2-4^2\)
\(=\left(x^2+x-1-4\right)\left(x^2+x-1+4\right)\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
(x2 - x)2 - 2 * (x2 - x) - 15
đặt x2 - x = a
có: a2 - 2a - 15 = (a2 - 2a + 1) - 16 = (a - 1)2 - 16 = (a - 5) (a + 3)
thay vào đc: (x2 - x - 5) (x2 - x +3)
(x^2+x)^2-2(x^2+x)-15
=(x2+x)2-2(x2+x)+1-16
=(x2+x-1)2-16
=(x2+x-1+4)(x2+x-1-4)
=(x2+x+3)(x2+x-5)
Ta có : \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15=\left[\left(x^2+x\right)-2\left(x^2+x\right)+1\right]-16=\left(x^2+x-1\right)^2-4^2\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
a) \(x^8+x^4-2\)
\(=x^8+x^7+x^6+x^5+2x^4+2x^3+2x^2+2x-x^7-x^6-x^5-x^4-2x^3-2x^2-2x-2\)
\(=x\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)-\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)
\(=\left(x-1\right)\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)
\(=\left(x-1\right)\left[x^4\left(x^3+x^2+x+1\right)+2\left(x^3+x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^4+2\right)\left(x^3+x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+2\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)\left(x+1\right)\)
c) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=x^4+2x^3+x^2-2x^2-2x-15\)
\(=x^4+2x^3-x^2-2x-15\)
\(=x^4+x^3+3x^2+x^3+x^2+3x-5x^2-5x-15\)
\(=x^2\left(x^2+x+3\right)+x\left(x^2+x+3\right)-5\left(x^2+x+3\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
( x2 + 4x )2 - 2( x2 + 4x ) - 15
= x4 + 16x2 - 2x2 - 8x - 15
= x2 + 14x2 - 6x - 15
(x2+x)2-2(x2+x)-15=
= (x2+x)2-2(x2+x) +1 -16
= [ ( x2 + x )- 1 ]2 - 42
= ( x2 + x - 1 )2 - 42
= ( x2 + x - 1 - 4 )( x2 + x - 1 + 4 )
= ( x2 + x - 5)( x2 + x + 3)