Lời giải:
c. 

$(x-3)(x^2+3x+9)-x^3=x^3-3^3-x^3=-27$ không phụ thuộc vào giá trị của biến

Ta có đpcm

d. 

$(3x+2)(9x^2-6x+4)-9x(3x^2+1)+9x$

$=(3x)^3+2^3-27x^3-9x+9x$

$=27x^3+8-27x^3=8$ không phụ thuộc vào giá trị của biến 

Ta có đpcm

c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x^3\)

\(=x^3-27-x^3\)

=-27

d) Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)+9x\)

\(=27x^3+8-27x^3-9x+9x\)

=8

( 3x - 2 )( 9x² + 6x + 4 ) - ( 2x - 5 )( 2x + 5 ) = ( 3x - 1 )³ - ( 2x + 3 )² + 9x( 3x - 1 )

⇔ 27x³ - 8 - ( 4x² - 25 ) = 27x³ - 27x² + 9x - 1 - ( 4x² + 12x + 9 ) + 27x² - 9x

⇔ 27x³ - 8 - 4x² + 25 = 27x³ - 1 - 4x² - 12x - 9

⇔ 27x³ - 4x² + 17 - 27x³ + 4x² + 12x + 10 = 0

⇔ 12x + 27 = 0

⇔ 12x = -27

⇔ x = -27/12 = -9/4

(3x-2) (9x+6x+4)-(3x-1) (9x+3x+1)=x-4

(3x - 2)(15x + 4) - (3x - 1)(12x + 1) = x - 4

<=> 45x² + 12x - 30x - 8 - (36x² + 3x - 12x - 1) - x + 4 = 0

<=> 9x² - 10x - 3 = 0

<=> (3x - \(\frac{5}{3}\))² = \(\frac{52}{9}\) => \(\orbr{\begin{cases}3x-\frac{5}{3}=\frac{2\sqrt{13}}{3}\\3x-\frac{5}{3}=-\frac{2\sqrt{13}}{3}\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{5+2\sqrt{13}}{9}\\x=\frac{5-2\sqrt{13}}{9}\end{cases}}\)

Vậy ...

1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

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

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

giúp mình với ạ câu nào cũng được

a: \(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)

\(=x^4-16-x^4+9=-7\)

b: \(=27x^3-8-27x^3+6=-2\)

c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)

Có ai làm dc câu này ko thầy mk cho đề hack não quá

\(\frac{12x^4-6x^3-9x^2}{-3x^2}-\left(2-3x\right)\left(2+3x\right)=-\left(3x+1\right)\)\(Dk:-3x^2\ne0\)\(< =>x\ne0\)

<=> \(-4x^2+2x+3-\left(2-3x\right).\left(2+3x\right)=-\left(3x+1\right)\)

<=> \(-4x^2+2x+3-4-6x+6x+9x^2=-3x-1\)

<=>\(5x^2+5x=0\)

<=> \(\orbr{\begin{cases}x=-1\left(n\right)\\x=0\left(l\right)\end{cases}}\)

thank