\(\frac{3}{5.11}+\frac{5}{11.21}+\frac{7}{21.35}+\frac{9}{35.53}=\frac{1}{2}\left(\frac{6}{5.11}+\frac{10}{11.21}+\frac{14}{21.35}+\frac{18}{35.53}\right)\)

\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{2}.\frac{48}{265}=\frac{24}{265}\)

Sửa lại :

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{2}.\frac{48}{265}=\frac{24}{265}\)

\(2.A=\frac{6}{5.11}+\frac{10}{11.21}+\frac{14}{21.35}+\frac{18}{35.53}\)

\(2.A=\frac{11-5}{5.11}+\frac{21-10}{11.21}+\frac{35-21}{21.35}+\frac{53-35}{35.53}=\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\)

\(2A=\frac{1}{5}-\frac{1}{53}=\frac{48}{265}\)

A = \(\frac{48}{265}:2=\frac{24}{265}\)

\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)

\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)

\(=\dfrac{1.3.5}{1.5.7}\)

giúp mk vs các bn

A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101

A = 2 - 2/101 = 200/101

B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51

B = 3-3/51(tự tính nhé)

C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31

C = 5(5-1/31)(tự tính)

D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)

2E nhân lên rồi giải giống trên

3F Rồi nhân 4/77 và rút gọn thì tính được

a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)

A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))

A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0

A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)

a) A= \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\) 

=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)

=\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right).\frac{3}{2}\)

=\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right).\frac{3}{2}\)

= \(\left(1-\frac{1}{50}\right).\frac{3}{2}=\frac{49}{50}.\frac{3}{2}=\frac{147}{100}\)

c) \(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

= \(\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right).5\)

= \(\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right).5\) 

= \(\left(1-\frac{1}{31}\right).5=\frac{30}{31}.5=\frac{150}{31}\)

Mấy bài còn lại mik đang phải nháp đã. Bạn thông cảm cho mik

9( 1/4.5 + 1/ 5.6 +.....+1/35.36)

=9 ( 1/4 - 1/5 +1/5 -1/6 +1/6 -1/7 +........+1/35-1/36 )

= 9(1/4 - 1/ 36)

=9.2/9=2

\(=\frac{219}{520}=\frac{155052}{368160}\)

\(=\frac{303}{708}=\frac{157560}{368160}\)

\(\frac{155052}{368160}< \frac{157560}{368160}\)

VẬY \(\frac{303}{708}\)LỚN HƠN

\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)

=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)

=\(\dfrac{450}{119}\)

Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)

\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)

A= 1/18 nhé bạn

xin lỗi vì mình không viết cách làm..........

toán hại não!!!!!!!!!!
 

A=\(\frac{2^{35}.9^{25}.5^{25}.13^{22}.7^{16}.5^{16}}{4^{17}.7^{17}.9^{26}.13^{22}.5^{22}.5^9.5^9}=\frac{2^{35}.5^1}{4^{17}.7^1.9}=\frac{2^{35}.5}{2^{34}.7^1.9}\)= \(\frac{2.5}{7.9}=\frac{10}{63}\)