giải phá các ngoặc ra rút gọn x rồi tìm

ko ghi lại đề nha ! 

a) \(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+x\left(2^2-x^2\right)=0\)

\(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+4x-x^3=0\)

\(\Leftrightarrow-27+4x=0\)

\(\Leftrightarrow4x=27\)

\(\Leftrightarrow x=6,75\) 

b)\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=-0,5\)

\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x+28=28\)

\(\Leftrightarrow3x^2+26x=0\)\(\Leftrightarrow x\left(3x+26\right)=0\)

Suy ra x=0 hoặc x=-26/3

cho mk hỏi ngu tí là 6x^2 ở đâu thế ạ

a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6                

=> (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6

=> (x² + 5x + 6) - (x² + 3x - 10) = 6                        

=> x² + 5x + 6 - x² - 3x + 10 = 6

=> 2x +16 = 6       => 2x = -10        => x = -5 

b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)     

  => (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6

=> (6x² + 31x + 18) - (6x² + 13x + 2) = 7             

=> 6x² + 31x + 18 - 6x² - 13x - 2 = 7

=> 18x + 16 = 7  => 18x = 9  => x = 0,5

c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0

=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0

=> 3(6x² - 5x +1) - (18x² - 29x + 3) = 0

=> (18x² -15x + 1) -(18x² - 29x +3) = 0

=> 18x² - 15x +1 -18x² + 29x - 3 = 0

=> 14x = 0  => x = 0

a)(x+2)(x+3)-(x-2)(x+5)=6

x(x+3)+2(x+3)-x(x+5)+2(x+5)=6

x²+3x+2x+6-x²-5x+2x+10=6

(x²-x²)+(3x+2x-5x+2x)+(10+6)=6

2x+16=6

2x=6-16

2x=-10

x=-10/2

x=-5. Vậy x=-5

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

6x²+27x+4x+18-6x²-x-12x-2=7

(6x²-6x²)+(27x+4x-x-12x)+(18-2)=7

18x+16=7

18x=7-16

x=-9/18=-1/2. Vậy x=-1/2

c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0

(9x-3)(2x-1)-(2x-3)(9x-1)=0

9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0

18x²-9x-6x+3-18x²+2x+27x-3=0

(18x²-18x²)+(27x+2x-6x-9x)+(3-3)=0

14x=0

x=0/14

x=0. Vậy x=0

a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6                    => (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6

=> (x² + 5x + 6) - (x² + 3x - 10) = 6                          

=> x² + 5x + 6 - x² - 3x + 10 = 6

=> 2x +16 = 6  => 2x = -10    => x = -5 

b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)

=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6

=> (6x² + 31x + 18) - (6x² + 13x + 2) = 7

=> 6x² + 31x + 18 - 6x² - 13x - 2 = 7

=> 18x + 16 = 7  => 18x = -9  => x = -0,5

c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0

=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0

=> 3(6x² - 5x + 1) - (18x² - 29x + 3) = 0

=> 18x² - 15x + 3 - 18x² + 29x -3 = 0

=> 14x = 0  => x = 0.

Cho g(x) = 0

x + 1 = 0

x = -1

Để f(x) chia hết cho g(x) thì x = -1 cũng là nghiệm của f(x)

Hay f(1) = 0

3.1² + 2.1² - 7.1 - m + 2 = 0

-2 - m + 2 = 0

m = 0

Vậy m = 0 thì f(x) chia hết cho g(x)

Giải chi tiết của em đây :

F(x) = 3x² + 2x² - 7x - m + 2 

F(x) \(⋮\) x + 1 \(\Leftrightarrow\) F(x) \(⋮\) x - (-1)

Theo bezout ta có : F(x) \(⋮\) x - (-1) \(\Leftrightarrow\) F(-1) = 0

\(\Leftrightarrow\) 3(-1)² + 2(-1)² - 7.(-1) - m + 2 = 0

    3 + 2 + 7 - m + 2 =0

              14 - m = 0

                     m = 14

Kết luận với m = 14 thì F(x) chia hết cho x + 1 

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}=3\)

b) \(x^2-4x=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

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

⇔ \(\left(x^3-3.x^2.3+3.x.3^2-3^3\right)-\left(x^3-3^3\right)+9x+9=15\)

⇔ \(x^3-9x^2+27x-27-x^3+27+9x+9=15\)

⇔ \(36x-9x^2+9=15\)

⇔ \(9x\left(4-x\right)=6\)