\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+\frac{1}{19.24}+...+\frac{1}{44.49}\right)\frac{1-3-7-...-49}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{1+\left(-3\right)+\left(-7\right)+...+\left(-49\right)}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{1-\left(3+7+...+49\right)}{89}\)

\(=\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\left(49+3\right).24:2\right)}{89}\)

\(=\frac{9}{196}.\frac{-623}{89}\)

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

\(=\frac{-9}{28}\)

CHÚC BN HỌC TỐT!!!!

b)=1/5.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/44-1/49).2-1-3-5-7-...-49/89

=1/5.(1/4-1/49).2-(1+3+5+7...+49)/89

=1/5.45/196.2-625/89

=9/196.-623/89

=9/196.-7

=9/28

h cho mình nha ! Chúc bạn học tốt

\(a,\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}=\frac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^3\cdot3^4\cdot3^6}=\frac{3^{10}\cdot2^3\cdot\left(3^2-2^3\right)}{2^3\cdot3^{10}}=3^2-2^3=1\)

\(b,\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)

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

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(3+49\right)\cdot24\div2}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{505}{89}\)

\(=\frac{1}{5}\cdot\frac{45}{196}\cdot\frac{505}{89}\)

bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó

#)Giải :

Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)

\(\Rightarrow5A=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\)

\(\Rightarrow5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)

\(\Rightarrow5A=\frac{1}{4}-\frac{1}{49}=\frac{45}{196}\)

\(\Rightarrow A=\frac{45}{196}\div5=\frac{9}{196}\)

Thay A vào B, ta được :

\(B=\frac{9}{196}.\frac{1-3-5-...-49}{89}\)

\(B=\frac{9}{196}.\frac{1-\left(3+5+7+...+49\right)}{89}\)

\(B=\frac{9}{196}.\frac{1-\left[\frac{\left(49+3\right).\left(\frac{49-3}{2}+1\right)}{2}\right]}{89}\)

\(B=\frac{9}{196}.\frac{-623}{89}=-\frac{9}{28}\)

B = \(\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right).\frac{1-3-5-...-49}{89}\)

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

    = \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(24.52:2\right)}{89}\)

    = \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-624}{89}\)

    = \(\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)

    = \(\frac{-9}{28}\)

Vậy B = \(-\frac{9}{28}\)

\(b.\)ghi lại đề nha bn

\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)

\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)

\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)

\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)

\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)

\(=\frac{2.2306}{1+1-\frac{2}{231}}\)

\(=\frac{2.2306}{2-\frac{2}{231}}\)

\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)

\(=\frac{2306}{1-\frac{1}{231}}\)

mình nha bn thanks nhìu <3

a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)

\(=\frac{1}{2017}\)

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