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b)3x^2-18x+27=3x^2-9x-9x+27=3x*(x-3)-9*(x-3)=(x-3)*(3x-9)=(x-3)*3*(x-3)=3*(x-3)^2
c)x^3-4x^2-12x+27=(x+3)*(x^2-3x+9-4)=(x+3)*(x^2-3x+5)
d)27x^3-1/27=(3x-1/3)*(9x^2-x+1/9) (hang dt)
con a) voi e) mk chiu
\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)
\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)
\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
a. x3 - 3x2 + 3x - 1
= (x-1)3
b. (x+y)2 - 4x2
=(x+y-4x)(x+y+4x)
c. 27x3 + 1/8
= (3x)3 +(1/2)3
=(3x+ 1/2) (9x - 3x.1/2 - 1/4)
d. ( x+y)3 - (x-y)3
= [(x+y)-(x-y)] [(x+y)2 + (x+y)(x-y) + (x-y)2]
=(x+y-x+y)[x2+2xy+y2+x2-y2+x2-2xy+y2)
=2y . (3x2+y2)
Mấy câu này ko biết đúng hay sai :{
\(3x^2-5x+2=3x^2-3x-2x+2=3x\left(x-1\right)-2\left(x-1\right)=\left(3x-2\right)\left(x-1\right)\)
b.)x^4+5x^3+15x-9
=x^4-9+5x^3+15x
=(x^2-3)(x^2+3)+5x(x^2+3)
=(x^2+3)(x^2-3+5x)
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
a, \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)
Mình có làm ở câu dưới rồi . Bạn tham khảo link :
https://olm.vn/hoi-dap/detail/231817932107.html
a) co sai de ko
b)x3-2x2+4x2-8x+3x-6=x2(x-2)+4x(x-2)+3(x-2)=(x-2)(x2+4x+3)=(x-2)(x+3)(x+1)
c)x3-2x2+2x2-4x-3x+6=x2(x-2)+2x(x-2)-3(x-2)=(x-2)(x2+2x-3)=(x-2)(x+3)(x-1)
d)x3-3x2+x2-3x-2x+6=x2(x-3)+x(x-3)-2(x-3)=(x-3)(x2+x-2)=(x-3)(x+2)(x-1)
a)\(\left(3x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(3x-\dfrac{1}{3}\right)\left(9x^2+x+\dfrac{1}{9}\right)\)
b)\(x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x^2+5x-24\right)\left(x-1\right)\)
\(=\left(x^2+8x-3x-24\right)\left(x-1\right)\)
\(=\left[x\left(x+8\right)-3\left(x+8\right)\right]\left(x-1\right)\)
\(=\left(x-3\right)\left(x+8\right)\left(x-1\right)\)
c)\(x^2+7x+12=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
d)\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)