bài 2 tính tổng
A=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\)
B=\(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{97.100}\)
C=\(\frac{8}{7.14}+\frac{8}{14.21}+......+\frac{8}{91.98}\)
giúp mk vs mk đang cần lắm
bài 2 tính tổng
A=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\)
B=\(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{97.100}\)
C=\(\frac{8}{7.14}+\frac{8}{14.21}+......+\frac{8}{91.98}\)
giúp mk vs mk đang cần lắm
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
a, \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
b,\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
c,\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)
a) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}\)
\(=\frac{2.1}{1}=2\)
b) \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5}-\frac{3}{10}\)
\(=\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)
\(=\frac{1}{3}+\frac{\frac{30}{60}+\frac{20}{60}-\frac{12}{60}}{\frac{45}{60}+\frac{30}{60}-\frac{9}{60}}\)
\(=\frac{1}{3}+\frac{\frac{19}{30}}{\frac{11}{10}}\)
\(=\frac{1}{3}+\frac{19}{33}=\frac{11}{33}+\frac{19}{33}\)
\(=\frac{30}{33}=\frac{10}{11}\)
tính các tổng sau
c=\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\)
d=1-2+3-4+...+99-100+101
ai bít giải giúp mink với
\(=\left(1+3+5+...+99+101\right)-\left(2+4+6+...98+100\right)\)
Thấy từ 1 đến 100 có (101-1)/2+1=51
=> 1+3+5+....+99+100=(1+101)x50/2=2601
Từ 2 đến 100 có (102-2)/2+1=50
=> 2+4+...+98+100=(2+100)X50/2=2550
=> D=2601-2550=51
2/2*4 + 2/4*6 + 3/6*8 + ... + 2/38*50
= 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + .... + 1/38 - 1/50
= 1/2 - 1/50
= 24/50
= 12/25
Mk có cách khác câu d nè
\(D=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)+101\)
\(=-1-1-1-...-1+101\)(có 50 số -1)
\(=-1\times50+101\)
\(=51\)
chúc bn hok tốt
Tính :
a)\(\left[-\frac{1}{2}\right]^2:\frac{1}{4}-2\left[-\frac{1}{2}\right]^2\)
b)\(-\frac{7}{37}.1\frac{3}{8}-4\frac{5}{8}.\frac{7}{12}+\frac{1}{2}\)
c)\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
d)\(H=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
BÀI TẬP: THỰC HIỆN PHÉP TÍNH;
a.\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{97.100}\) b.\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
c.\(\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{\frac{49}{11}-\frac{7}{8}-\frac{35}{12}}\) d.\(\frac{5-\frac{5}{3}-\frac{5}{9}-\frac{5}{27}}{-8+\frac{8}{3}+\frac{8}{9}+\frac{8}{27}}\div\frac{15-\frac{15}{11}-\frac{15}{121}}{16-\frac{16}{11}-\frac{16}{121}}\)
e.\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)}\) f.\(\frac{3}{5}-\frac{2}{3}\left(1-0,3\right)-\left(-3\right)^2\)
a) đặt B = \(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{97.100}\)
\(B=5.\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)\)
đặt A = \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\)
A = \(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+...+\frac{1}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)\)
A = \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
A = \(\frac{1}{3}.\left(1-\frac{1}{100}\right)\)
A = \(\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(\Rightarrow B=5.\frac{33}{100}=\frac{33}{20}\)
b) đặt C = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
3C = \(3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
3C - C = \(\left(3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
2C = \(3-\frac{1}{3^{100}}\)
\(\Rightarrow\)C = \(\frac{3-\frac{1}{3^{100}}}{2}\)
c) \(\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{\frac{49}{11}+\frac{7}{8}-\frac{35}{12}}=\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{7.\left(\frac{7}{11}+\frac{1}{8}-\frac{5}{12}\right)}=\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{7.\left(\frac{5}{12}+\frac{1}{8}-\frac{7}{11}\right)}=\frac{1}{7}\)
e) \(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)}=\frac{2^3.5^3.7^4}{2.5.7^2}=2^2.5^2.7^2=70^2\)
f) \(\frac{3}{5}-\frac{2}{3}\left(1-0,3\right)-\left(-3\right)^2\)
\(=\frac{3}{5}-\frac{2}{3}.\frac{7}{10}+9=\frac{3}{5}-\frac{7}{15}+9=\frac{2}{15}+9=\frac{137}{15}\)
Tính tổng: \(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
Tính tổng: \(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
\(\frac{x}{2^2}\)+\(\frac{x}{2^3}\) +\(\frac{x}{2^4}\) =\(\frac{x}{3^2}\) +\(\frac{x}{3^3}\) +\(\frac{x}{3^4}\) là x =
Tính tổng:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....+\frac{1}{97.100}=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.......+\frac{1}{97}-\frac{1}{100}\right)=\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
Gọi dãy phân số trên là A
A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
A = \(1-\frac{1}{100}\)
A = \(\frac{99}{100}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
=\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
=\(\frac{1}{3}.\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{3}.\frac{99}{100}\)=\(\frac{33}{100}\)
Tính:
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)