\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{17}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\dfrac{9}{20}\cdot x=\dfrac{45}{23}\)
\(x=\dfrac{45}{23}:\dfrac{9}{20}\)
\(x=\dfrac{100}{23}\)

\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{1}{17.20}\right)x=\dfrac{45}{23}\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)

\(\Rightarrow\dfrac{9}{20}.x=\dfrac{45}{23}\)

\(\Rightarrow x=\dfrac{100}{23}\)

3/2.5 + ...+ 3/17 .20

= 3/2 .(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 120)

= 3/2 . (1/2 - 1/20)

= \(\frac{3}{2}\) . \(\frac{9}{20}\) = \(\frac{27}{40}\)

P= 3/ 2.5 + 3/5.8 + 3/8.11 + .... + 3/17.20

P= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + .... + 1 /17 - 1/20

P= 1 / 2 - 1 / 20

P = 9/20

\(P=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(\frac{9}{20}\)

\(A=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

\(A=\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+...+\frac{3}{17}+\frac{3}{20}\)

\(A=\frac{3}{2}-\frac{3}{20}=\frac{30}{20}-\frac{3}{20}=\frac{27}{20}\)

vậy A \(=\frac{27}{20}\)

mình nghĩ là 9/20

A=1/2 -1/5 +1/5 -1/8+...+1/17 -1/20=1/2 -1/20=9/20

yeah!

mình làm đúng rồi

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

a, A = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\)

A = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...\dfrac{1}{17}-\dfrac{1}{20}\)

A = \(\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{10}{20}-\dfrac{1}{20}=\dfrac{9}{20}\)

Lần sau nếu có bài dạng như thế này bạn hãy làm theo quy tắc sau nha: \(\dfrac{m}{b.\left(b+m\right)}=\dfrac{m}{b}-\dfrac{m}{b+m}\)

Tick cho mk vs. thank!!!

b, B = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)

B = \(\dfrac{5^2}{5}.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)

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

B = \(5.\left(\dfrac{1}{1}-\dfrac{1}{31}\right)\)

B = \(5.\dfrac{30}{31}=\dfrac{150}{31}\)

Tick cho mk nha! please đó!

Sao chưa có ai trả lời vậy

đặt A=1/2.5 +1/5.8 + 1/811+...+1/17.20

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(3A=\frac{1}{2}-\frac{1}{20}\)

\(A=\frac{9}{20}:3\)

\(A=\frac{3}{20}\)

Sai đề => Sửa: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{20}\)

\(\Rightarrow\frac{9}{20}\)

nói j chứ trả lời sao đc

dễ ẹt tí nữa làm cho

a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}\)

\(=\dfrac{9}{20}\)

b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{4}-\dfrac{1}{10}\)

\(=\dfrac{3}{20}\)

c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)

\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)

\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)

\(=4\left(1-\dfrac{1}{49}\right)\)

\(=4.\dfrac{48}{49}\)

\(=\dfrac{192}{49}\)