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

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

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

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

C= 3/2.5 + 3/5.8 + ... + 3/17.20

C = 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20

C = 1/2 - 1/20

C = 9/20

Mẹo để nhận diện khi gặp dạng này là hai số ở mẫu trừ nhau thì ra được số ở tử.

Ta có :
\(\frac{3}{(k) * (k+3)} = \frac{1}{k} - \frac{1}{k+3}\)
Áp dụng vào biểu thức C, ta được:
C = \(\frac{3}{2 * 5} + \frac{3}{5 * 8} + \frac{3}{8 * 11} + ... + \frac{3}{17 * 20}\) = \(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + ... + \frac{1}{17} - \frac{1}{20}\)
= \(\frac{1}{3} - \frac{1}{20}\) = \(\frac{17}{60}\)
Vậy C = \(\frac{17}{60}\)

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=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