\(\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{6}{18}+\dfrac{10}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)

\(\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}\)

\(3:\dfrac{5}{9}=3.\dfrac{9}{5}=\dfrac{27}{5}\)

\(3.\dfrac{5}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{1}{9}.\dfrac{9}{3}=\dfrac{1}{3}\)

\(\dfrac{1}{3}:\dfrac{1}{7}=\dfrac{7}{3}\)

\(9+\dfrac{9}{3}=9+3=12\)

\(4-\dfrac{2}{4}=4-\dfrac{1}{2}=\dfrac{7}{2}\)

\(\dfrac{1}{3}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{9}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3+5}{9}\) \(=\) \(\dfrac{8}{9}\)

\(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) \(=\) \(\dfrac{9}{63}\) \(-\) \(\dfrac{7}{63}\) \(=\) \(\dfrac{9-7}{63}\) \(=\) \(\dfrac{2}{63}\)

\(\dfrac{1}{9}\) \(\times\) \(\dfrac{9}{3}\) \(=\) \(\dfrac{1\times9}{9\times3}\) \(=\) \(\dfrac{1}{3}\)

\(\dfrac{1}{3}\) \(\div\) \(\dfrac{1}{7}\) \(=\) \(\dfrac{1}{3}\) \(\times\) \(\dfrac{7}{1}\) \(=\) \(\dfrac{1\times7}{3\times1}\) \(=\) \(\dfrac{7}{3}\)

\(3\) \(\div\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{1}\) \(\div\)  \(\dfrac{5}{9}\) \(=\dfrac{3}{1}\times\dfrac{9}{5}=\dfrac{3\times9}{1\times5}=\dfrac{27}{5}\)

\(3\times\dfrac{5}{9}=\dfrac{3}{1}\times\dfrac{5}{9}=\dfrac{3\times5}{1\times9}=\dfrac{5}{3}\)

\(9+\dfrac{9}{3}=\dfrac{9}{1}+\dfrac{9}{3}=\dfrac{27}{3}+\dfrac{9}{3}=\dfrac{27+9}{3}=\dfrac{36}{3}=12\)

\(4\) \(-\dfrac{2}{4}=\dfrac{4}{1}-\dfrac{2}{4}=\dfrac{16}{4}-\dfrac{2}{4}=\dfrac{14}{4}=\dfrac{7}{2}\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

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

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

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

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

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

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

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

Tính:

2/7 + 5/7=1          

8/15 + 6/15 =14/15         

2/9 + 3/9 + 4/9=9/9=1

[ Không cần làm phép tính ]Điền vào ô trống <,>,= :

3/5 + 1/5 = 1/5 + 3/5

7/9 + 1/9 < 7/9 + 1/9 + 1/9

3/7 + 2/7 < 3/7 + 4/7

1/5 + 3/5 < 2/5 + 3/5

Tính:

2/7 + 5/7=1          

8/15 + 6/15 =14/15         

2/9 + 3/9 + 4/9=9/9=1

[ Không cần làm phép tính ]Điền vào ô trống <,>,= :

3/5 + 1/5 = 1/5 + 3/5

7/9 + 1/9 < 7/9 + 1/9 + 1/9

3/7 + 2/7 < 3/7 + 4/7

1/5 + 3/5 < 2/5 + 3/5

1 +1 +5 +6 + 9+ 2+ 3+ 5+ 6+ 5+ 9+ 5+ 3+ 2+ 8+ 1+ 5+ 5+ 9+ 9+ 8+ 8+ 8+ 8+ 8+ 8+ 6+ 3+ 3+ 3+ 2+ 1+ 1+ 1+5 +4 + 5+ 2+ 2

31 + 25+ 63+ 96+ 5+ 3+ 65+ 66+ 666+ 6+ 56+ 2+ 2+45 +5 + 5+ 456 +49 + 49+496+ 4554 +45  +562 +45 +693 +42+ 258 +6 +3 +3 +3 + 3+ 3+ 3+ 3+ 3+ 3+ 3+6 +6 + 6+ 6 +6 +6 + 6+ 6 +9 +9 + +9+ 9+ 9+9+6+9+9+9+9+9+9+9+9+9+9+9+9+9+9+9+9+9+9+  9 + 9+ 9+9 + 9+ 9+ 9+ 9+ 9+ 9+ 9+ 9+9+9 + 9+9+ 9+ 9+ 9+9+9+ 9+ 9 + 9 +9 +9+9 + 9+ 9+9+ 9 + 9+ 9+ 9+ 9+ 9+ 9+ 9 +9  +  9  +9 +9 +9 + 9+ 9 +9 +9 + 9 +9 + 9+ 9+ 9 (+ =+ 9 +9 +9 +9 +9 + 9+ 9 + 9 + 9 + 9+ 9 + 9 + 9 + 9 +9 + 9+9 +9+9+ 9+ 9+ 9 +9++ 9 +9 + 9+9+9+9+9 ++9 + 9+9 + =)

chỗ này là sao +=+9+...(ở trên cộng nhiều rồi +=+ 9 là sai đề luôn rồi )

Bài làm

\(=\left(\varnothing\right)\)

# Chúc bạn học tốt #

bằng 1038373 nhé

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)

mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)

\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)

\(\Rightarrow A< B\)

Ta thấy : A= ( 1+5+5^2+.......+5^9)/(1+5+5^2+...... +5^8)= 5^9

B=(1+3+3^2+......+3^9)/(1+3+3^2+,,,,,,,,+3/9)=1

mÀ 5^9 > 1 . SUY RA  A>B

Vậy A>B

mk ko chắc chắn lắm 

k cho mk nhé

a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)

\(=\dfrac{9}{16}\)

a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)

\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)

\(=1-\dfrac{29}{15}\)

\(=\dfrac{15-29}{15}\)

\(=-\dfrac{14}{15}\) 

b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)

\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=\dfrac{7}{9}\cdot1\)

\(=\dfrac{7}{9}\)

c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\) 

\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)

\(=\dfrac{25}{63}\)

d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)

\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)

\(=3\cdot\dfrac{9}{11}\)

\(=\dfrac{27}{11}\)