C= 3/2.5 + 3/5.8 + ... + 3/17.20
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S
7 tháng 4 2023
\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{17}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\dfrac{9}{20}\cdot x=\dfrac{45}{23}\)
\(x=\dfrac{45}{23}:\dfrac{9}{20}\)
\(x=\dfrac{100}{23}\)
TA
7 tháng 4 2023
\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{1}{17.20}\right)x=\dfrac{45}{23}\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)
\(\Rightarrow\dfrac{9}{20}.x=\dfrac{45}{23}\)
\(\Rightarrow x=\dfrac{100}{23}\)
NT
4
19 tháng 3 2017
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)
= \(\frac{3}{2}\) . \(\frac{9}{20}\) = \(\frac{27}{40}\)
C
30 tháng 10 2016
\(A=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
\(A=\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+...+\frac{3}{17}+\frac{3}{20}\)
\(A=\frac{3}{2}-\frac{3}{20}=\frac{30}{20}-\frac{3}{20}=\frac{27}{20}\)
vậy A \(=\frac{27}{20}\)
V
11
DH
0
HH
18 tháng 3 2017
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!!!
HH
18 tháng 3 2017
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!
TN
11 tháng 5 2016
đặt A=1/2.5 +1/5.8 + 1/811+...+1/17.20
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(A=\frac{9}{20}:3\)
\(A=\frac{3}{20}\)
MN
6 tháng 4 2017
Sai đề => Sửa: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{20}\)
\(\Rightarrow\frac{9}{20}\)
9 tháng 3 2018
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. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)
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 = \(\frac{17}{60}\)