\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2012.2014}\)

\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2012}-\frac{1}{2014}\right)\)

\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2014}\right)\)

\(\Leftrightarrow A=\frac{1}{2}\cdot\frac{503}{1007}\)

\(\Leftrightarrow A=\frac{503}{2014}\)

= 1/2[1/2 - 1/4+1/4-1/6 + 1/6-1/8+...+ 1/2012-1/2014]

= 1/2[1/2-1/2014]

= 1/2 * 503/1007

= 503/2014

Ta có:

|7-2x|+7=2x

|7-2x|    =2x-7

=>7-2x=2x-7 hoặc 7-2x=-(2x-7)

Với 7-2x=2x-7                                  Với 7-2x=-(2x-7)

=>7+7=2x+2x                                     =>7-2x=-2x+7

   14=4x                                             =>7-7=-2x+2x      

=>4x=14                                            =>0=0x

=>x=\(\frac{7}{2}\)                                           =>x=0  

Vậy x=\(\frac{7}{2}\);x=0

x chỉ có thể bằng 7/2 chứ không thể bằng 0

\(4x^2-9=0\)
\(4x^2=9\)
\(x^2=\frac{9}{4}\)
\(x^2=\frac{3}{2}^2\)
\(x=\frac{3}{2}\)

4x² - 9 = 0

4x² = 0 + 9

4x² = 9

( 2x )² = 3²

=> 2x = 3

x = \(\frac{3}{2}\)

c) 3x² - 10x + 7 \(\ge\)0

<=> 3x² - 3x - 7x + 7 \(\ge\)0

<=> 3x(x - 1) - 7(x-1) \(\ge\)0

<=> (x-1)(3x - 7) \(\ge\)0

<=> x - 1 \(\ge\) 0 hoặc 3x - 7 \(\ge\)0

<=> x  \(\ge\) 1 hoặc x  \(\ge\)7/3

  Vậy: ......

d) 4x² + 9x + 5 \(\le\)0

<=>4x² + 4x + 5x + 5 \(\le\)0

<=>4x(x + 1) + 5(x + 1) \(\le\)0

<=>(x + 1)(4x + 5) \(\le\)0

<=>x + 1 \(\le\)0 hoặc 4x + 5 \(\le\)0

<=>x  \(\le\)-1 hoặc x  \(\le\)-5/4

Nguyễn Duyên 

bấm vào đây là có lời giải cụ thể https://h.vn/hoi-dap/question/402297.html