phân tích đa thức thành nhân tử
\(8x^3+60x^2y+150xy^2+125y^3\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(8x^3+60x^2y+150xy^2+125y^3=\left(2x\right)^3+3.\left(2x\right)^2.5y+3.2x.\left(5y\right)^2+\left(5y\right)^3\)
\(=\left(2x+5y\right)^3\)
a) = (3x)^2 + 2.3x.5+ 5^2 = (3x+5)^2
b) = (2/3x)^2-(4y)^2=(2/3x-4y)(2/3x+4y)
c) = -(9x^4-12/5x^2y^2+4/25y^4) = -[(3x^2)^2 - 2.3x^2.2/5y^2 + (2/5y^2)^2]= -(3x^2-2/5y^2)^2
d) = (x-5)^2 - 4^2= (x-5+4)(x-5-4) = (x-1)(x-9)
e) = (2x)^3 + 3.(2x)^2.(5y) + 3.(2x).(5y)^2 + (5y)^3 = (2x+5y)^3
f) = (8x)^2 - (8a+b)^2 = (8x-8a-b)(8x+8a+b)
g) = (7x-4-2x-1)(7x-4+2x+1) = (5x-5)(9x-3) = 5(x-1).3(x-3)=15(x-1)(x-3)
h) = (x-y)(x+y)- 2(x+y) = (x+y)(x-y-2)
# Chúc bạn học tốt #
a
\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)
b
\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)
a: 8x^3-1/125y^3
=(2x)^3-(1/5y)^3
=(2x-1/5y)(4x^2+2/5xy+1/25y^2)
b: =(2y-x)^3
a) Ta có : 64x2 - (8x + y)2
= (8x)2 - (8x + y)2
= (8x - 8x - y) (8x + 8x + y)
= -y(16x + y)
Bài 1:
a)\(64x^2-(8x+y)^2=-y\left(16x+y\right)\)
b)\((x+y+15)^2-2(x+y+15)+1\)
\(=\left(x+y+14\right)^2\)
c)\(8x^3+60x^2y+150xy^2+125y^3\)
\(=\left(2x+5y\right)^3\)
d)\(x^{16}-1\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
Bài 2:
\(\left(n+7\right)^2-\left(n-5\right)^2\)
\(=\left(n+7+n+5\right)\left(n+7-\left(n+7\right)\right)\)
\(=24\left(n+1\right)\) chia hết 24
a: \(20m^3n^2-15m^4n^2+25m^5n^2\)
\(=5m^3n^2\left(4-3m+5m^2\right)\)
c: \(2x^4-32\)
\(=2\left(x^4-16\right)\)
\(=2\left(x^2+4\right)\left(x^2-4\right)\)
\(=2\left(x^2+4\right)\left(x-2\right)\left(x+2\right)\)
\(8x^3+60x^2y+150xy^2+125y^3\)
\(=\left(8x^3+125y^3\right)+\left(60x^2y+150xy^2\right)\)
\(=\left(2x+5y\right)\left(4x^2-10xy+25y^2\right)+30xy\left(2x+5y\right)\)
\(=\left(2x+5y\right)\left(4x^2-10xy+25y^2+30xy\right)\)
\(=\left(2x+5y\right)\left(4x^2+20xy+25y^2\right)\)
\(=\left(2x+5y\right)\left(2x+5y\right)^2=\left(2x+5y\right)^3\)
C2 : Áp dụng luôn HĐT ( x + y ) 3
\(8x^3+60x^2y+150xy^2+125y^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.5y+3.2x.\left(5y\right)^2+\left(5y\right)^3\)
\(=\left(2x+5y\right)^2\)
=.= hok tốt!!