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

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

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

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

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

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

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

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

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

2 x 3 x 12 + 4 x 6 x 42 + 8 x 27 x 3

= 2 x 12 x 3 + 4 x 6 x 42 + 8 x 3 x 27

= 24 x 3 + 24 x 42 + 24 x 27

= 24 x ( 3 + 42 + 27 )

= 24 x 72

= 1728

Nho tick cho minh nhe ban😁

tk mk , mk tk 

          (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 $x=0$

3.

ĐKXĐ: $x\geq -1$

PT \(\Leftrightarrow (1+\sqrt{x^2+4033}).\frac{(x+2016)-(x+1)}{\sqrt{x+2016}+\sqrt{x+1}}=2015\)

\(\Leftrightarrow 1+\sqrt{x^2+4033}=\sqrt{x+2016}+\sqrt{x+1}\)

\(\Leftrightarrow (1+\sqrt{x^2+4033})^2=(\sqrt{x+2016}+\sqrt{x+1})^2\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VP}\leq 2(x+2016+x+1)=4x+4034\)

\(\text{VP}=x^2+4034+2\sqrt{x^2+4033}\geq x^2+4034+2\sqrt{4033}>x^2+4034+5\)

Mà: $x^2+4034+5-(4x+4034)=(x-2)^2+1> 0$

$\Rightarrow x^2+4034+5> 4x+4034$

$\Rightarrow \text{VP}> \text{VT}$

Do đó pt vô nghiệm.

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