Bài 1. Đề khó nhìn quá mình không làm được ._.

Bài 2.

12x² + 24x - 15 = ( 2x - a )( 6x - 3 )

<=> 12x² + 24x - 15 = 12x² - 6x - 6ax + 3a

<=> 12x² + 24x - 15 = 12x² + ( -6 - 6a )x + 3a

Đồng nhất hệ số ta được :

\(\hept{\begin{cases}-6-6a=24\\-15=3a\end{cases}}\Leftrightarrow a=-5\)

Bài 3: 

=>-3<x<2

Bài 5.5:

\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)

\(\Leftrightarrow\left(2x^2+2x-3x-3\right)+2x\cdot\left(2x^2-3x-3\right):\left(-2x\right)=18\)

\(\Leftrightarrow2x^2-x-3-2x^2+3x+3=18\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=\dfrac{18}{2}\)

\(\Leftrightarrow x=9\) 

B1:

\(x^n\left(x+1\right)-x^n-x^{n-1}=0\)

\(\Rightarrow x^{n-1}\left(x^2+x\right)-x^{n-1}.x-x^{n-1}=0\)

\(\Rightarrow x^{n-1}\left(x^2+x-x-1\right)=0\)

\(\Rightarrow x^{n-1}\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^{n-1}=0\\x^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)

b: \(D=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x-x^2+1}{3x}\)

\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

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

\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{-4x+2}+\dfrac{x^2-3x-1}{3x}\)

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

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

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

c: Khi x=1 thì \(D=\dfrac{1-1}{3}=0\)

1: =>(x+2018)(6x-3)=0

=>x+2018=0 hoặc 6x-3=0

=>x=1/2 hoặc x=-2018

2: x(x-11)+3(11-x)=0

=>(x-11)(x-3)=0

=>x=11 hoặc x=3

4: =>(x+5)(2x-4)=0

=>2x-4=0 hoặc x+5=0

=>x=2 hoặc x=-5

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

=>x=3 hoặc x=-2

Bài 1:

\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)

\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)

\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 2:

\(x\left(x-11\right)+3\left(11-x\right)=0\)

\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)

Câu 3:

\(x\left(x-3\right)-2\left(3-x\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Câu 4:

\(2x\left(x+5\right)-4\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Bài 1:

a: \(2x^2-8x=0\)

=>\(x^2-4x=0\)

=>x(x-4)=0

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

b: \(\left(x+2\right)^2-x\left(x-1\right)=10\)

=>\(x^2+4x+4-x^2+x=10\)

=>5x+4=10

=>5x=6

=>\(x=\dfrac{6}{5}\)

c: \(x^3-6x^2+9x=0\)

=>\(x\left(x^2-6x+9\right)=0\)

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

=>\(\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

 Bài 1:

a; (\(x+1\)).(\(x+2\)) - (\(x-1\)).(\(x-5\)) = 0

    \(x^2\) + 2\(x\) + \(x+2\) - \(x^2\) + 5\(x\) + \(x\) - 5 = 0

   (\(x^2\) - \(x^2\)) + (2\(x\) + \(x+5x+x\))- (5  -2) = 0

        0 + (3\(x\) + 5\(x\) + \(x\)) + 0 - 3 = 0

                 8\(x\) + \(x\) - 3 = 0

                 9\(x\) = 3

                    \(x=\dfrac{3}{9}\)

Vậy \(x=\dfrac{1}{3}\)

b; (2\(x\) - 1)² + 4.(5 - \(x\)) = 15

     4\(x^2\) - 4\(x\) + 1 + 20 - 4\(x\) = 15

     4\(x^2\) - (4\(x\) + 4\(x\)) + (1 + 20 - 15) = 0

        4\(x^2\) - 8\(x\) + 6 = 0

         4.(\(x^2\) - 2\(x\) + 1) + 2 = 0

         4(\(x-1\))² + 2 = 0

Vì 4.(\(x-1\))² ≥ 0 ⇒ 4.(\(x-1\))² + 2  ≥ 3 > 0 (\(\forall x\))

Vậy không có giá trị nào của \(x\) thỏa mãn đề bài

Kết luận \(x\) \(\in\) \(\varnothing\)

Bài 2:

a; M = \(x^2\) - 6\(x\) + 6 

    M = (\(x^2\) - 2.3\(x\) + 3²) - 3

   M = (\(x\) - 3)² - 3 vì  (\(x-3\))² ≥ 0 ∀ \(x\); ⇒ (\(x-3\))²  - 3 ≥ -3

Vậy M_min = - 3 khi \(x-3\) = 0 ⇒ \(x=3\) 

Vậy Giá trị nhỏ nhất của biểu thức là -3 khi \(x=3\)