Tính tổng sau:
S = \(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+...+\(\dfrac{1}{300}\)
Tính nhanh
a, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) + \(\dfrac{1}{99}\) + \(\dfrac{1}{143}\)
b, A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\)
c, H =\(\dfrac{4047991-2010x2009}{4050000-2011x2009}\)
d, T = \(\dfrac{2009x20010+2000}{2011x2010-2020}\)
e, P = \(\dfrac{7589-80,5x69,3}{7485,05-79x69,3}\)
f, B = 5,1 x 42,2 + 1,7 x 448 x 3 - 0,15 x 700
Giúp mình với
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
Tính tổng : \(1-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{3}-\dfrac{1}{28}-\dfrac{1}{6}-\dfrac{1}{21}\)
\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)
\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)
\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)
Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)
a) Tìm các số nguyên n để phân số sau có giá trị nguyên:
\(A=\dfrac{n-5}{n-3}\)
\(\dfrac{n+4}{n+1}\)
b) Tính A, biết A= \(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\).
Nếu \(n-3=1\Rightarrow n=4\); \(n-3=-1\Rightarrow n=2\); \(n-3=2\Rightarrow n=5\); \(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)
\(A=\dfrac{n+4}{n+1}\) làm tương tự.
b) Dễ thấy các số ở mẫu có thể viết dưới dạng:
\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)
\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)
\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)
...
\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)
Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=\dfrac{3}{8}\)
tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
ăn lồn cái địt mẹ mày lửa trùa ĐẦU LỒN nhá
find the value of:
\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+...+\(\dfrac{1}{300}\)
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)
\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)
\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)\)
\(=2\cdot\dfrac{23}{50}=\dfrac{23}{25}\)
Vậy A = \(\dfrac{23}{25}\).
It's very easy :)))))
Call expression is A, we have:
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}.\)
\(\Rightarrow2A=2\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\right).\)
\(\Rightarrow2A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}.\)
\(\Rightarrow2A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}.\)
\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\right).\)
\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right).\)
\(\Rightarrow2A=2\left(\dfrac{25}{50}-\dfrac{2}{50}\right).\)
\(\Rightarrow2A=2.\dfrac{23}{50}=\dfrac{23}{25}.\)
Vậy.....
Câu 1: Tính giá trị biểu thức:
a.A=\(\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\).\(\dfrac{21}{24}\)
b.B=\(\dfrac{5}{6}\)+6\(\dfrac{5}{6}\)\(\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right)\):8\(\dfrac{1}{3}\)
c.C=1+3+6+10+15+...+1225.
A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}\)
Ta có:
\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)
\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)
\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)
\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)
\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)
\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)
\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)
\(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)
\(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)
\(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)
\(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)
\(=\dfrac{28}{36}=\dfrac{7}{9}\)
Vậy: \(A=\dfrac{7}{9}\)
D = 1 - \(\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{3}-\dfrac{1}{28}-\dfrac{1}{6}-\dfrac{1}{21}\)
\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)
\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)
Vậy D=1/4
\(\dfrac{y}{2022}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)