a/ \(2x^2+8x+1=2\left(x^2+4x+\frac{1}{2}\right)=2\left(x^2+2.2x+4-4+\frac{1}{2}\right)\)

     \(=2\left[\left(x+2\right)^2-\frac{7}{2}\right]=2\left(x+2\right)^2-7\ge-7\)

       Vậy Min A = -7 khi x + 2 = 0 => x = 2

b/ \(2x^2+3x+1=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=2\left(x^2+2.\frac{3}{4}.x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}\right)\)

          \(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

         Vậy Min B = -1/8 khi x + 3/4 = 0 => x = -3/4

\(B=\left|y-2018\right|+\left|2017-y\right|>=\left|-2018+2017\right|=1\)

Dấu '=' xảy ra khi (y-2018)(y-2017)<=0

=>2017<=y<=2018

a) \(\frac{3}{4}-\frac{2}{5}.x=x\)

\(\Rightarrow\frac{-2}{5}.x-x=\frac{-3}{4}\)

\(x.\left(\frac{-2}{5}-1\right)=\frac{-3}{4}\)

\(x.\frac{-7}{5}=\frac{-3}{4}\)

\(x=\frac{-3}{4}:\left(\frac{-7}{5}\right)\)

\(x=\frac{15}{28}\)

b) (2x-1).(3x-1/5).(4-2x) = 0

=> 2x - 1 = 0 => 2x = 1 => x = 1/2

3x-1/5 = 0 => 3x = 1/5 => x = 1/15

4-2x = 0 => 2x = 4 => x = 2

KL: x = 1/2 hoặc x = 1/15 hoặc x = 2

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

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

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

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

b: Để A>0 thì x-3>0

hay x>3

nguyenthihuyentrang

( 2x  - 5)²  - ( 4x - 1 ) ( x + 3 ) = 5 

=> ( 2x ) ²   - 2 . 2x. 5 + 5² - 4x² + 12x - 3 - x = 5 

=> 4x²    - 20x + 15 - 4x²  + 11x - 3 = 5 

=> -20x + 11x = 5 + 3 - 15 

=> -9x = -7  =>  x = 7/9 

^^ Học tốt! 

M.n giup mk vs ak!! mk se tag k cho m.n, camon ah!

ê sao ko cho 

\(trog.khug.ý\)