- Nhẩm tính rồi điền kết quả vào chỗ trống.

- Biểu thức có hai phép tính thì thực hiện từ trái sang phải.

1 + 2 = 3     1 + 1 = 2     1 + 2 = 3     1 + 1 + 1 = 3

1 + 3 = 4     2 - 1 = 1     3 - 1 = 2     3 - 1 - 1 = 1

1 + 4 = 5     2 + 1 = 3     3 - 2 = 1     3 - 1 + 1 = 3

tính;

1+2=3   1+1=2  1+2=3  1+1+1=3

1+3=4   2-1=1  3-1 =2   3-1-1= 0

1+4=5   2+1=3  3-2=1   3-1+1=3

2 - 1 = 1     3 - 1 = 2     1 + 1 = 2     1 + 2 = 3

3 - 1 = 2     3 - 2 = 1     2 - 1 = 1     3 - 2 = 1

3 - 2 = 1     2 - 1 = 1     3 - 1 = 2     3 - 1 = 2

2 - 1 = 1     3 - 1 = 2     1 + 1 = 2     1 + 2 = 3

3 - 1 = 2     3 - 2 = 1     2 - 1 = 1     3 - 2 = 1

3 - 2 = 1     2 - 1 = 1     3 - 1 = 2     3 - 1 = 2

ok nhá

Lời giải chi tiết:

#HT#

a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)

b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)

=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).

d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).

e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)

Lời giải chi tiết:

4333344

?reeeeeeeeeeee

a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(A=1-\frac{1}{2^{50}}

Bạn Detective_conan giải đúng đấy!

2 câu đầu thôi bạn ak

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+4+...+2018}\right)\)

\(A=\frac{2}{1+2}\cdot\frac{2+3}{1+2+3}\cdot\frac{2+3+4}{1+2+3+4}\cdot...\cdot\frac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018}\)

Đến chỗ này đố ai tính được ?!!?!

gạch các số của tử số và các số của mẫu số giống nhau

ví dụ như bạn nói:

 \(\dfrac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018} =1\)