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\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
đặt A = 1/4.9 + 1/9.14+ 1/14.19 + .....1/44.49
ta có 5.A = 5/4.9 + 5/9.14+ 5/14.19 + .....5/44.49 = 1/4- 1//9 + 1/9 - 1/14+........+ 1/44 -1/49 = 1/4 - 149 = 45/196
suy ra A = 9/196
đặt B = 1-3--5-...-49 = 1 - (3+5+ ....+ 49)
đặt C = 3+5+...+49 khoảng cách là d = 2
số các số hạng là (49-3)/2 + 1 = 24
tổng C = (49+3)/2 x 24 = 624
suy ra B = 1-624 = -623
vậy A = 9/196 .(-623)/89 = -9/28
\(A=9-\frac{3}{5}+\frac{2}{3}-7-\frac{7}{5}+\frac{3}{2}-3+\frac{9}{5}-\frac{5}{2}\)
\(=\left(9-7-3\right)+\left(\frac{9}{5}-\frac{7}{5}-\frac{3}{5}\right)+\left(\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2-\frac{1}{5}=-\frac{11}{5}\)
\(\left(-\frac{5}{12}\right):\frac{7}{3}-\left(-\frac{5}{12}\right):\frac{7}{4}=\left(-\frac{5}{12}\right):\left(\frac{7}{3}-\frac{7}{4}\right)=\left(-\frac{5}{12}\right):\frac{7}{12}=-\frac{5}{7}\)
\(\left[\left(\frac{2}{5}\right)^0\right].\frac{19}{13}-\left(\frac{7}{3}\right)^{2019}.\frac{3}{7}^{2019}\)
\(=\left(\frac{2}{5}\right)^0.\frac{19}{13}-\left(\frac{7}{3}.\frac{3}{7}\right)^{2019}\)
\(=1.\frac{19}{13}-1^{2019}\)
\(=1.\frac{19}{13}-1\)
\(=\frac{19}{13}-1\)
\(=\frac{6}{13}\)
Bài giải
a, \(\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)
\(=\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)
\(=\left(-\frac{5}{12}\right)\cdot\frac{3}{7}-\left(-\frac{5}{12}\right)\cdot\frac{4}{7}\)
\(=\frac{-15}{84}+\frac{20}{84}=\frac{5}{84}\)
b, \(\left[\left(\frac{2}{5}\right)^0\right]^{2020}\cdot\frac{19}{37}-\left(\frac{7}{3}\right)^{2019}\cdot\frac{3^{2019}}{7}\)
\(=1^{2020}\cdot\frac{19}{37}-\frac{7^{2019}}{3^{2019}}\cdot\frac{3^{2019}}{7}\)
\(=\frac{19}{37}-7^{2018}\)
Đặt \(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-3-5-7-...-49}{89}\)
Đặt \(B=1-3-5-7-..-49\)
\(=1-\left(3+5+7+...+49\right)\)
\(=1-\left\{\left(49+3\right).\left[\left(49-3\right):2+1\right]:2\right\}\)
\(=1-624\)
\(=-623\)
\(\Rightarrow\dfrac{9}{196}.\left(\dfrac{-623}{89}\right)=-\dfrac{9}{28}\)
Vậy: \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}=-\dfrac{9}{28}\)
Xét \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}.\dfrac{45}{196}\)
=\(\dfrac{9}{196}\)
Xét \(\dfrac{1-3-5-7-..-49}{89}\)
=\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)
CT tính sl số hạng (số cuối - số đầu ):2+1
số lượng số hạn của dãy 3+5+7+...+49 là (49-3):2+1=24
Áp dụng CT tính tổng số hạng dãy số cách đều Tổng = [ (số đầu + số cuối) x Số lượng số hạng ] : 2
=> tổng = [(3+49).24]:2=624
=>\(\dfrac{1-624}{89}\)
=\(\dfrac{-623}{89}\)
=-7
từ đó ta có \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)
Đặt \(A=1+7+7^2+...7^{50}\)
\(7\cdot A=7+7^2+7^3+.....+7^{51}\)
\(7\cdot A-A=\left(7+7^2+7^3+.....+7^{51}\right)-\left(1+7+7^2+....+7^{50}\right)\)
\(A.\left(7-1\right)=\left(7-7\right)+\left(7^2-7^2\right)+.....+\left(7^{50}-7^{50}\right)+7^{51}-1\)
\(A\cdot6=7^{51}-1\Rightarrow A=\frac{7^{51}-1}{6}\)