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\(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)
b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)
c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
a) \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x+8y\right)\left(\frac{1}{5}x-8y\right)\)
b) \(x^3+\frac{1}{27}=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
c) \(-x^3+9x^2-27x+27\)
\(=27-x^3+9x^2-27x\)
\(=\left(3-x\right)\left(9+3x+x^2\right)+9x\left(x-3\right)\)
\(=\left(3-x\right)\left(9+3x+x^2\right)-9x\left(3-x\right)\)
\(=\left(3-x\right)\left(9+3x+x^2-9x\right)\)
\(=\left(3-x\right)\left(9-6x+x^2\right)=\left(3-x\right)\left(9-3x-3x+x^2\right)\)
\(=\left(3-x\right)\left[3\left(3-x\right)-x\left(3-x\right)\right]=\left(3-x\right)\left(3-x\right)\left(3-x\right)=\left(3-x\right)^3\)
(Nhớ k cho mình với nha!, Mình chắc chắn là mình làm đứng luôn đó! Chúc may mắn nhá!)
a/ Ta có: \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
b/ \(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
c/ Đề sai
a)27x2.(y-1)9x3.(1-y)
=27x2.(y-1)+9x3.(y-1)
=9x2(y-1)[3+x]
b)8x3 + 1/27
=(2x)3 + (\(\frac{1}{3}\))3
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
a) 27x2 ( y - 1) - 9x3 ( 1 - y)
=27x2 (y-1) + 9x3 ( y - 1 )
= (27x2 + 9x3) ( y -1 )
=9x2 ( x + 3) ( y - 1)
b)8x3+1/27
\(=\left(2x\right)^3+\left(\frac{1}{3}\right)^3\)
\(=\left(\frac{2x}{9}+\frac{1}{27}\right)\left(36x^2-6x+1\right)\)
c)49 ( y - 4 )2 - 9 ( y + 2)2
= [7(y - 4)]2 - [3(y + 2)]2
= (7y - 28 + 3y + 6)(7y - 28 - 3y - 6)
= (10y - 22)(4y - 34)
= 4(5y - 11)(2y - 34)
\(x^3+\frac{1}{x^3}=x^3+\left(\frac{1}{x}\right)^3=\left(x+\frac{1}{x}\right)\left(x^2-x+\frac{1}{x^2}\right)\)( x khác 0 )
\(-x^3+9x^2-27x+27=-\left(x^3-9x^2+27x-27\right)=-\left(x-3\right)^3\)
\(\left(xy+1\right)^2-\left(x-y\right)^2=\left(xy+1-x+y\right)\left(xy+1+x-y\right)\)
a) 1 - 2y + y2
= (1-y)2
b) ( x + 1 )2 - 25
=( x + 1 )2 - 52
=(x+1+5)(x+1-5)
27x3 + 27x2 + 9x + 1 + x + 1/3
= ( 27x3 + 27x2 + 9x + 1 ) + 1/3( 3x + 1 )
= ( 3x + 1 )3 + 1/3( 3x + 1 )
= ( 3x + 1 )[ ( 3x + 1 )2 + 1/3 ]
= ( 3x + 1 )( 9x2 + 6x + 1 + 1/3 )
= ( 3x + 1 )( 9x2 + 6x + 4/3 )
\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)
\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)
\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
-x3+9x2-27x+27
-x3+3x2+6x2-18x-9x+27
-x2(x-3)+6x(x-3)-9(x-3)
(-x2+6x-9)(x-3)
-(x2-6x+9)(x-3)
-(x-3)3
a,\(0,04x^2-64y^2=\left(0,2x\right)^2-\left(8y\right)^2=\left(0,2x-8y\right)\left(0,2x+8y\right)\)
b,\(-x^3+9x^2-27x+27=-x^3+3x^2+6x^2-18x-9x+27\)
\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)=\left(-x^2+6x-9\right)\left(x-3\right)\)
\(=-\left(x-3\right)^3\)
Nhớ tick mình nha bạn,cảm ơn nhiều nha.