Ta có: \(A=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{49-4}{4\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{-623}{89}=-\dfrac{9}{28}\)

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 \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^3+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^6.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}\left(3^5-3^4\right)}{2^6\left(3^6-3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9\left(7^3+2^3.7^3\right)}\)

\(A=\frac{2^6\left(3^5-3^4\right)}{3^6-3^5}-\frac{5\left(7^3-7^4\right)}{7^3+2^3.7^3}\) ( \(\frac{5\left(7^3-7^4\right)}{7^3.1+2^3.7^3}=\frac{5\left(7^3-7^4\right)}{7^3\left(1+2^3\right)}\))

\(A=\frac{2^6\left(3^5-3^4\right)}{3^6-3^5}-\frac{5\left(7^3-7^4\right)}{7^3\left(1+2^3\right)}\)

\(A=\frac{2^6.162}{486}-\frac{5.\left(-2058\right)}{7^3.9}\)

\(\Rightarrow A=\frac{648}{486}-\frac{-12540}{3087}\)

bạn tự tính nha máy tính mình mất r

\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)

\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

= \(\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)

= \(\dfrac{2}{3.4}-\dfrac{5\left(-6\right)}{9}\)

= \(\dfrac{7}{2}\)

\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)

Ta có: �=(14⋅9+19⋅14+114⋅19+...+144⋅49)⋅1−3−5−7−...−4989A=(4⋅91​+9⋅141​+14⋅191​+...+44⋅491​)⋅891−3−5−7−...−49​

⇔�=15⋅(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−3−5−7−...−4989⇔A=51​⋅(4⋅95​+9⋅145​+14⋅195​+...+44⋅495​)⋅891−3−5−7−...−49​

⇔�=15⋅(14−19+19−114+114−119+...+144−149)⋅1−3−5−7−...−4989⇔A=51​⋅(41​−91​+91​−141​+141​−191​+...+441​−491​)⋅891−3−5−7−...−49​

⇔�=15⋅(14−149)⋅1−3−5−7−...−4989⇔A=51​⋅(41​−491​)⋅891−3−5−7−...−49​

⇔�=15⋅(49−44⋅49)⋅1−3−5−7−...−4989⇔A=51​⋅(4⋅4949−4​)⋅891−3−5−7−...−49​

⇔�=15⋅45196⋅1−3−5−7−...−4989⇔A=51​⋅19645​⋅891−3−5−7−...−49​

⇔�=9196⋅1−3−5−7−...−4989⇔A=1969​⋅891−3−5−7−...−49​

⇔�=9196⋅−62389=−928⇔A=1969​⋅89−623​=−289​
 

A = -9/28

Bài 1 :

a) +) \(\dfrac{1}{8}\cdot16^n=2^n\)

\(\Leftrightarrow\dfrac{1}{8}=\dfrac{2^n}{16^n}\)

\(\Rightarrow\dfrac{1}{8}=\dfrac{1}{8}^n\)

Vậy n = 1.

+) \(27< 3^n< 243\)

\(\Leftrightarrow3^3< 3^n< 3^5\)

Vậy n = 4.

Bài 2 : \(\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}=-\dfrac{45}{28}\)

=\(\dfrac{1}{5}\).(\(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+....+\dfrac{5}{44.49}\)).\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)

=\(\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\).\(\dfrac{1-624}{89}\)

=\(\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\).(-7)

=\(\dfrac{1}{5}\).\(\dfrac{45}{196}\).(-7)=\(\dfrac{-9}{28}\)

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