\(...\Rightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)

\(\Rightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)

\(\Rightarrow45x+9=15\Rightarrow45x=6\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)

`9x^2+6x-8=0`

`<=> 9x^2+12x-6x-8=0`

`<=> 3x(3x+4) - 2(3x+4)=0`

`<=>(3x+4)(3x-2)=0`

`<=> 3x+4=0` hoặc `3x-2=0`

`<=> 3x=-4` hoặc `3x=2`

`<=>x=-4/3` hoặc `x=2/3`

__

`2x^2 +3x-27=0`

`<=> 2x^2+9x-6x-27=0`

`<=>x(2x+9) - 3(2x+9)=0`

`<=> (2x+9)(x-3)=0`

`<=> 2x+9=0` hoặc `x-3=0`

`<=> 2x=-9` hoặc `x=3`

`<=>x=-9/2` hoặc `x=3`

\(\dfrac{3-3x}{\left(1+x\right)^2}:\dfrac{6x^2-6}{x+1}\)

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

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

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

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

b) Bạn có thể viết kiểu latex được không ạ ?

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

c: \(\Leftrightarrow2x+1=3\)

hay x=1

`4(x-6)-x^2 (2+3x)+x(5x-4)+3x^2 (x-1)`

`=4x-24-2x^2 -3x^3 +5x^2-4x+3x^3-3x^2`

`=-24`

\(4\left(x-6\right)-2x\left(2+3x\right)+x\left(5x-4\right)+3x2\left(x-1\right)\\ =4x-24-4x-6x^2+5x^2-4x+6x^2+6x\\ =2x+5x^2-24\)

\(B=x^2-x^3+x^3+27=x^2+27\)

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)-3x\left(1-x\right)\)

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

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

\(=24x-6x^2\)

Hình như đề có chỗ sai sót ở đâu đó bạn .

vãi ò ông ngx thành đạt chép sai đầu bài r (x-1)³ cchchuchưchứchứ kkoko pphphaphaiphảiphải (x-3)³³

Ta có: \(\left(1-x\right)^2+\left(x-x^2\right)+3=0\)

\(\Leftrightarrow x^2-2x+1+x-x^2+3=0\)

\(\Leftrightarrow4-x=0\)

hay x=4

Vậy: S={4}

$⇔x^2-2x+1+x-x^2+3=0$

$⇔-x=-4$

$⇔x=4$

Vậy phương trình đã cho có tập nghiệm S={4}