Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
3x^2-8x+4
=3x^2-6x-2x+4
=3x(x-2)-2(x-2)
=(x-2)(3x-2)
l-kie cho mik nha bạn
a) \(=3\left(x^2-10x+25\right)=3\left(x-5\right)^2\)
b) \(=x\left(x+y\right)+8\left(x+y\right)=\left(x+y\right)\left(x+8\right)\)
c) \(=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
\(\text{a) }4x^{16}+81=4x^4+36x^2+81-36x^8\)
\(=\left(4x^{16}+36x^8+81\right)-36x^8\)
\(=\left[\left(2x^8\right)^2+2.2x^8.9+9^2\right]+\left(6x^4\right)^2\)
\(=\left(2x^8+9\right)^2-\left(6x^4\right)^2\)
\(=\left(2x^8+9-6x^4\right)\left(2x^8+9+6x^4\right)\)
\(\text{b) }x^4+2018x^2+2017x+2018\)
\(=x^4+2018x^2+2018x-x+2018\)
\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)
\(=x\left(x^3-1\right)-2018\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2-x\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)
a) \(3x^2+8x-11\)
\(=3x^2-3+11x-11\)
\(=\left(3x^2-3x\right)+\left(11x-11\right)\)
\(=3x\left(x-1\right)+11\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+11\right)\)
b) \(x^4+2018x^2-2017x+2018\)
\(=\left(x^4+x\right)+\left(2018x^2-2018x+2018\right)\)
\(=x\left(x^3+1\right)+2018\left(x^2-x+1\right)\)
\(=x\left(x+1\right)\left(x^2-x+1\right)+2018\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left[x\left(x+1\right)+2018\right]\)
\(=\left(x^2-x+1\right)\left(x^2+x+2018\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+2018\right)\)
a) 3x2 + 8x - 11
=3x2+11x-3x-11
=x(3x+11)-(3x+11)
= (x-1)(3x+11)