\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{2014}-\frac{1}{2016}\)

\(=\frac{1}{2}-\frac{1}{2016}\)

\(=\frac{1008}{2016}-\frac{1}{2016}=\frac{1007}{2016}\)

\(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{2014\times2016}\)

=\(\left(\frac{2}{2}-\frac{2}{4}\right)+\left(\frac{2}{4}-\frac{2}{6}\right)+...+\left(\frac{2}{2014}-\frac{2}{2016}\right)\)

=\(\frac{2}{2}-\frac{2}{4}+\frac{2}{4}-\frac{2}{6}+...+\frac{2}{2014}-\frac{2}{2016}\)

=\(\frac{2}{2}-\frac{2}{2016}=\frac{1007}{1008}\)

tôi không biết

`#3107.101107`

A,

\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)

B,

\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)

Help

tk mk , mk tk 

Câu a:

C = 1 + 5^2 + 5^4 + 5^6 + ... + 5^ 2018

5^2C= 5^2 + 5^4 + ...+ 5^ 2020

25C - C = 5^2 + 5^4 + ...+ 5^ 2020 - ( 1 + 5^2 + ... + 5^ 2018)

24C = (5^2 - 5^2) +...+(5^2018 - 5^2018) + (5^2020 - 1)

24C = 0 + 0 +...+ 5^2020 - 1

C = (5^2020 - 1)/24

Câu b:

D = 2 x 4 + 4 x 6 + 6 x 8 +...+ 2016 x 2018

6D = 2.4.6 + 4.6.6 + ... + 2016.2018.6

6D = 2.4.6 + 4.6.(8 - 2) + 6.8.(10 - 4) +...+2016.2018.(2020-2014)

6D = 2.4.6 + 4.6.8 - 2.4.6 + 6.8.10 - 4.6.8 +...+2016.2018.2020 - 2014.2016.2018

6D = 2016.2018.2020

D= 2016.2018.2020: 6

D = (2016 : 6).(2018.2020)

D = 336.4076360

D = 136956960

          (1+2+3+4+5+6+7+8+9+...............................+2016+2025) x (24,2 - 24,2)                                                                                   = (1 + 2 +3+4+5+6+7+8+9+...............................+2016+2025) x 0                                                                                                       = 0                                                                                          

1.

ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$

PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)

\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)

\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)

\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)

\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)

2.

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$

$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$

$\Leftrightarrow 2\sqrt{x}=0$

$\Leftrightarrow x=0$

Thử lại thấy thỏa mãn 

Vậy $x=0$

a) 612 : (x - 7) + 8 = 59

   612 : (x - 7) = 51

     x - 7 = 612 : 51 = 12

      x = 12 + 7

      x = 19

b) 2016 - ( 2+4+6+8....+x)= 376

    2 + 4 + 6+ 8 + ... +x = 2016 - 376 = 1640

    Áp dụng công thức tính dãy số ta có :

\(\frac{\left[\left(x-2\right):2+1\right].\left(x+2\right)}{2}=1640\)

\(\Rightarrow\left(\frac{x-2}{2}+1\right).\left(x+2\right)=3280\)

\(\Rightarrow\frac{x}{2}.\left(x+2\right)=3280\)

\(\Rightarrow\frac{x^2}{2}+x=3280\)

\(\Rightarrow\frac{x^2}{2}+\frac{2x}{2}=3280\)

\(\Rightarrow x^2+2x=6560\)

=> x . (x+2) = 6560

=> x = 80