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ta tách 2/5x7 = 2/5-2/7 tách những cái kia tương tự góp vào rồi tính
Đặt P= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)
Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)
Từ (1) và (2) suy ra:
A= 5:\(\dfrac{1}{7}\)=5.7=35
Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 35
a) \(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{1}{100}+x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1+\dfrac{1}{100}\)
\(\Rightarrow x=\dfrac{500}{300}-\dfrac{300}{300}+\dfrac{3}{300}\)
\(\Rightarrow x=\dfrac{203}{300}\)
b) \(\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{19.24}\right)-x+\dfrac{1}{3}=\dfrac{7}{3}\)
=>\(\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{19}-\dfrac{1}{24}\right)-x=\dfrac{7}{3}-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{11}-\dfrac{1}{24}-x=2\)
\(\Rightarrow-x=2-\dfrac{1}{11}+\dfrac{1}{24}\)
\(\Rightarrow-x=\dfrac{528}{264}-\dfrac{24}{264}+\dfrac{11}{264}\)
\(\Rightarrow x=\dfrac{515}{264}\)
c) Câu hỏi của Đàm Chu Hữu An - Toán lớp 6 - Học toán với OnlineMath
\(K=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)
\(=\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{89}\right)=\dfrac{5}{4}\cdot\dfrac{86}{267}=\dfrac{215}{534}\)
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)
Mk thực sự nghĩ đề hình như bị sai hay sao ấy!
D = \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{2006.2009}\)
= \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{2006}-\dfrac{1}{2009}\)
= \(\dfrac{1}{5}-\dfrac{1}{9}=\dfrac{2004}{10045}\)
C = \(\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\)
= \(\dfrac{10}{5}\left(\dfrac{5}{7.12}+\dfrac{5}{12.17}+\dfrac{5}{17.22}+...+\dfrac{5}{502.507}\right)\)
= \(\dfrac{10}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+....+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
= \(\dfrac{10}{5}\left(\dfrac{1}{5}-\dfrac{1}{507}\right)\)
= \(\dfrac{10}{5}.\dfrac{502}{2535}\)
= \(\dfrac{1000}{3549}\)
a: =-10/21
b: =-5/27
c: =-20/20=-1
d: =-15/105=-1/7
e: =-63/63=-1
\(\dfrac{10}{21}\)
\(\dfrac{5}{27}\)
\(1\)
\(\dfrac{1}{7}\)
\(1\)
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=2.\dfrac{3}{16}\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
\(\dfrac{10}{3\cdot7}-\dfrac{5}{7\cdot12}-\dfrac{7}{12\cdot19}-\dfrac{5}{19\cdot24}\)
\(=\dfrac{10}{3\cdot7}-\left(\dfrac{1}{7}-\dfrac{1}{12}\right)-\left(\dfrac{1}{12}-\dfrac{1}{19}\right)-\left(\dfrac{1}{19}-\dfrac{1}{24}\right)\)
\(=\dfrac{10}{21}-\dfrac{1}{7}+\dfrac{1}{12}-\dfrac{1}{12}+\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{1}{24}\)
\(=\dfrac{10}{21}-\dfrac{3}{21}+\dfrac{1}{24}\)
\(=\dfrac{7}{21}+\dfrac{1}{24}=\dfrac{1}{3}+\dfrac{1}{24}=\dfrac{8}{24}+\dfrac{1}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)