Viết các biểu thức sau dưới dạng tích của các đa thức:
a) 8x3-1000 b) 0,001+64x3 c) \(\dfrac{1}{125}\)y3+x3
d) 27x3-\(\dfrac{1}{8}\)y3 e) (x-1)3+27 f) \(\dfrac{x^6}{8}\)-y6
Giải chi tiết giúp mình nha.Cảm ơn.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
a) \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
b) \(125x^6-27x^9=\left(5x^2-3x^3\right)\left(25x^4+15x^5+9x^6\right)\)
c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+10x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\left(\dfrac{1}{5}y+x\right)\left(\dfrac{1}{25}y^2-\dfrac{1}{5}xy+x^2\right)\)
a, 8x3- 1000 = (2x)3 - 103 = (2x -10). (4x2 + 20x +100)
b,\(0,001+64x^3=\left(\dfrac{1}{10}\right)^3+\left(4x\right)^3=\left(\dfrac{1}{10}+4x\right).\left(\dfrac{1}{100}-\dfrac{2}{5}x+16x^2\right)\)
c, \(\dfrac{1}{125}y^3+x^3=\left(\dfrac{1}{5}y\right)^3+x^3=\left(\dfrac{1}{5}y+x\right).\left(\dfrac{1}{25}y^2-\dfrac{1}{5}yx+x^2\right)\)
\(d,27x^3-\dfrac{1}{8}y^3=\left(3x\right)^3-\left(\dfrac{1}{2}y\right)^3=\left(3x-\dfrac{1}{2}y\right).\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\)
a: \(1-\dfrac{x^3}{8}=\left(1-\dfrac{1}{2}x\right)\left(1+\dfrac{1}{2}x+\dfrac{1}{4}x^2\right)\)
b: \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)
c: \(64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
\(27-x^3\)
\(=3^3-x^3\)
\(=\left(3-x\right)\left(9+3x+x^2\right)\)
\(8x^3+0,001\)
\(=\left(2x\right)^3+\left(\dfrac{1}{10}\right)^3\)
\(=\left(2x+\dfrac{1}{10}\right)\left(4x^2-2x\dfrac{1}{10}+\left(\dfrac{1}{10}\right)^2\right)\)
\(=2\left(x+\dfrac{1}{5}\right)\left(4x^2-\dfrac{1}{5}x+\dfrac{1}{100}\right)\)
\(\dfrac{x^3}{125}-\dfrac{y^3}{27}\)
\(=\left(\dfrac{x}{5}\right)^3-\left(\dfrac{y}{3}\right)^3\)
\(=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left[\left(\dfrac{x}{5}\right)^2+\dfrac{x}{5}.\dfrac{y}{3}+\left(\dfrac{y}{3}\right)^2\right]\)
\(=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
b )
Dấu = thứ 3 :
Sửa lại : \(2\left(x+\dfrac{1}{20}\right)\)
\(a,27-x^3\)
\(=3^3-x^3\)
\(=\left(3-x\right)\left(9+3x+x^2\right)\)
Các câu còn lại lm tương tự nhé.
hok tốt!
a) \(27-x^3=\left(3-x\right)\left(9+3x+x^2\right)\)
b) \(8x^3+0,001=\left(2x+0,1\right)\left(4x^2-0,2x+0,01\right)\)
c) \(\frac{x^3}{125}-\frac{y^3}{27}=\left(\frac{x}{5}-\frac{y}{3}\right)\left(\frac{x^2}{25}+\frac{xy}{15}+\frac{y^2}{9}\right)\)
p/s: chúc bạn học tốt
\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)