a: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

=>\(\left(x+3\right)\cdot\left(27x^2-12x\right)=0\)

=>3x(x+3)(9x-4)=0

=>x(x+3)(9x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)

b: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)

=>\(\left(x-2\right)\left(3x+5\right)=2\left(x-2\right)\left(x+1\right)\)

=>(x-2)(3x+5)-(x-2)(2x+2)=0

=>(x-2)(3x+5-2x-2)=0

=>(x-2)(x+3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

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

=>\(2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

=>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)

nhận thấy x = 0 không là nghiệm của phương trình

Chia 2 vế phương trình cho x², ta được : 

\(x^2-9x+24-\frac{27}{x}+\frac{9}{x^2}=0\)  ( 1 )

đặt \(t=x+\frac{3}{x}\)

( 1 ) \(\Leftrightarrow\left(x+\frac{3}{x}\right)^2-9\left(x+\frac{3}{x}\right)+18=0\)

\(\Leftrightarrow t^2-9t+18=0\Leftrightarrow\left(t-6\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=6\\t=3\end{cases}}\)

Khi đó : \(\orbr{\begin{cases}x+\frac{3}{x}=6\Leftrightarrow x=3\pm\sqrt{6}\\x+\frac{3}{x}=3\Leftrightarrow x\in\varnothing\end{cases}}\)

a/. x³ - 9x² +27x - 19 = 0

<=> (x³ - 3.x² .3 + 3.3² .x - 3³) + 8 = 0

<=> (x - 3)³ + 8 = 0

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

<=> (x -1)(x² - 6x+ 9 -2x +6 +4) =0

<=> (x - 1)(x²  - 8x + 19) = 0

<=> x - 1 = 0 => x = 1

Vậy S = {1}

Xem lại đề câu b nha bạn?

c/. x³ + 1 -7x -7 =0 

<=> (x³ + 1) -7(x+1)=0

<=> (x+1)(x²-x+1) -7(x+1)=0

<=> (x+1)(x²-x+1-7)=0

<=> x + 1 = 0 hay x² -x - 6 = 0

<=> x = -1 hay (x² - 3x) + (2x - 6) = 0 

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

<=>                 (x - 3)(x+2) = 0

<=> x = -1 hay x = 3 hay x = -2

Vậy S = {-1;3;-2}

X^{3 -} X²-8X²+8X+19X-19=0

<=>X²(X-1)-8X(X-1)+19(X-1)=0

<=>(X-1)(X²-8X+19)=0

vi X^2-8X+19=(X-4)²+3>3

a, x³-x²- 8x² +8x+19x-19=0
<=> (x-1)(x²- 8x+19)=0
<=> x-1=0 <=> x=1
 

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)

=>\(-2\sqrt{3x}=-4\)

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

=>\(\sqrt{1-3x}=2\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

`<=>\sqrt{3x}=1`

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

`<=>\sqrt{x+1}=3`

`<=>x+1=9`

`<=>x=8` (t/m)