C

\(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{9x^2+6x+4}=3x-2\)

Giải:

\(\left(27x^3-8\right):\left(6x+9x^2+4\right)\)

\(=\dfrac{27x^3-8}{6x+9x^2+4}\)

\(=\dfrac{\left(3x-2\right)\left(9x^2+6x+4\right)}{6x+9x^2+4}\)

\(=3x-2\)

Vậy ...

khong biet

( 27\(x^3\) - 8) : ( 9\(x^2\) + 6x + 4)

= [ \(\left(3x\right)^3\) - 23] : ( 9\(x^2\) + 6x + 4)

= (3x - 2)( 9\(x^2\) + 6x + 4) : ( 9\(x^2\) + 6x + 4)

= 3x - 2

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

\(a,=\left(x+3\right)^3\\ b,=-\left(x-2\right)^3\\ c,=\left(\dfrac{x}{2}+y^2\right)^3\\ d,=\left(x-y-5\right)^3\)

b1 A=10000 B=23386

b2 a,x=5 

b,x=4 x=2 

có lẽ bn nên tự lm thì hơn

a: \(x^2-10x=-25\)

\(\Leftrightarrow x-5=0\)

hay x=5

b: \(x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)