​​4/15+ 4/35+ 4/63+ 4/99=16/33

​chúc bạn hok tốt

\(\frac{4}{12}+\frac{4}{35}+\frac{4}{63}+\frac{4}{99}\)

\(=2.\left(\frac{2}{12}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)

\(=2.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

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

\(=2.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=2.\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=2.\frac{8}{33}\)

\(=\frac{16}{33}\)

Tham khảo nhé~

\(\frac{4}{15}+\frac{4}{35}+\frac{4}{63}+\frac{4}{99}\)

\(=4\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=4\left(\frac{2}{21}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=4\left(\frac{1}{9}+\frac{1}{99}\right)\)

\(=4\cdot\frac{4}{33}=\frac{16}{33}\)

4/15 + 4/35 + 4/63 + 4/99 + 4/143

= 8/21 + 8/77 + 4/143

= 16/33 + 4/143

= 20/39

\(\frac{4}{15}+\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}\)

\(=2\times\left(\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{11\times13}\right)\)

\(=2\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=2\times\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=2\times\frac{10}{39}\)

\(=\frac{20}{39}\)

1+(-2)+3+(-4)+.......+19+(-20)

=(1+(-2))+(3+(-4))+....+(19+(-20)) có 10 nhóm như vậy

=(-1)+(-1)+.....+(-1)

=-10

a) 1 + (-2) + 3 + (-4) + ... + 19 + (-20)

= 1 - 2 + 3 - 4 + ... + 19 - 20

= ( 1 + 3 + ... + 19 ) - ( 2 + 4 + ... + 20 )

Số số hạng VT : ( 19 - 1 ) : 2 + 1 = 10 ( số )

Tổng VT = ( 19 + 1 ) . 10 : 2 = 100

Số số hạng VP : ( 20 - 2 ) : 2 + 1 = 10 ( số )

Tổng VP là : ( 20 + 2 ) x 10 : 2 = 110 

Ta có biểu thức :

100 - 110

= -10

b) 1 - 2 + 3 - 4 + ... + 99 - 100

= ( 1 + 3 + ... + 99 ) - ( 2 + 4 + ... + 100 )

Tương tự câu a) ta có biểu thức :

2500 - 2550

= -50

1-2-3+4+5-....+96+97-98-99+100

=(1-2-3+4)+....+(97-98-99+100)

=0+.........+0

=0

1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ... + 96 + 97 - 98 - 99 + 100

Ta thấy tổng trên có 100 số hạng. Ta chia tổng thành tường nhóm, mỗi nhóm có 4 số hạng như sau:

1 - 2 - 3 + 4 + 5 - 6 - 7 - 8 + ... + 96 + 97 - 98 - 99 + 100

= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ... + ( 97 - 98 - 99 + 100 )

= 0 + 0 + ... + 0

= 0

\(B=\dfrac{4}{3}+\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{143}\) 

    \(=4(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{143})\)

        vì \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{143}<\dfrac{1}{2}\) nên \(4(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{143})<4*\dfrac{1}{2}=2\Rightarrow B<2\)

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{63}+\frac{100}{99}\\ =\frac{2.2}{1.3}+\frac{4.4}{3.5}+\frac{6.6}{5.7}+\frac{8.8}{7.9}+\frac{10.10}{9.11}\)

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{65}+\frac{100}{99}\)

\(1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+1+\frac{1}{65}+1+\frac{1}{99}\)

\(\left(1+1+1+1+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{65}+\frac{1}{99}\right)\)

\(\frac{60}{11}\)

B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301

B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301

\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);

\(\frac{2}{15}=\frac{1}{3}-\frac{1}{5}\);

\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).

=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)

=> B=\(\frac{1}{1}-\frac{1}{301}\)

=> B=\(\frac{300}{301}\)

ko hỉu

de thiu em a

BÀI 1:

a)    \(\frac{4}{5}+\frac{3}{8}+\frac{1}{4}\)

\(=\frac{32}{40}+\frac{15}{40}+\frac{10}{40}\)

\(=\frac{32+15+10}{40}\)\(=\frac{57}{40}\)

b)    \(\frac{5}{9}+\frac{2}{3}+\frac{1}{2}\)

\(=\frac{10}{18}+\frac{12}{18}+\frac{9}{18}\)

\(=\frac{10+12+9}{18}=\frac{31}{18}\)

bài 23,, 5 nữa

1. 

a.57/40

b.31/18