\(E=\dfrac{3}{1x5}+\dfrac{3}{5x9}+...+\dfrac{3}{121x125}\)

\(\dfrac{4}{3}xE=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{121}-\dfrac{1}{125}\)

\(\dfrac{4}{3}xE=1-\dfrac{1}{125}\)

\(E=\dfrac{124}{125}x\dfrac{3}{4}=\dfrac{93}{125}\)

0,2x-2/3(x+1)=1/3

<=>0,2x-2/3x-2/3=1/3

<=>-7/15x=1

<=>x=-15/7

Vậy..............

Hok tốt

\(0,2x-\frac{2}{3}\left(x+1\right)=\frac{1}{3}\)

\(\frac{1}{5}x-\frac{2}{3}x+\frac{2}{3}=\frac{1}{3}\)

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

\(-\frac{7}{15}x=-\frac{1}{3}\)

\(x=\frac{5}{7}\)

day so tren co so so hang la

(2n-2):2+1=n(so hang)

tong tren la 

(2n+2).n:2=n.(n+1)

hok tot

2+4+6+8+..+2n

=(2n+2).[ ( 2n-2) : 2+1] :2

=2.(n+1).n:2

=n.(n+1)

Dãy số trên có số số hạng là: (khoảng cách mỗi số là $1$ đơn vị)

$(2020-1):1+1=2020$(số hạng)

Tổng của dãy số trên là:

$(2020+1)\times2020:2=2041210$

1+2+3+...+2020=\(\dfrac{\left(2020-1\right):1+1\cdot\left(1+2020\right)}{2}\)=2041210.

-15 +40 = 40-15=25

-100-(-100) =100 -100=0

(-5).75+(-5).25=(-5).(75+25)= - 500

3³ +2016^o.(-5) = 27+1.(-5) = 22

tinh nhanh;

(-27)+4²+18+27 = 2(8+9) =34

(-10-20).(-5) = (-30).(-5)    =150

Ta có:

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{15}=\frac{14}{15}\)

`2/5 - 3/8 + 7/6 : 7/5`

`= 2/5 - 3/8 + 7/6 xx 5/7`

`= 2/5 - 3/8 +5/6`

`= 1/40+5/6`

`=103/120`

\(=\frac{1}{3}+\frac{1}{15}+\frac{3}{5}-\left(\frac{3}{4}+\frac{2}{9}+\frac{1}{36}\right)+\frac{1}{64}=\frac{5+1+9}{15}-\frac{27+8+1}{36}+\frac{1}{64}.\)

\(=\frac{1}{64}\)

=1/64