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\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)
\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)
\(=-\frac{9}{10}+\frac{1}{90}\)
= ...
bn tự tính nha!
A = \(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)
=> 4A = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}\)
=> 3A = \(1-\frac{1}{4^{2012}}\)
=> A = \(\frac{1-\frac{1}{4^{2012}}}{3}\)
Vậy A \(< \frac{1}{3}\)
\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{3}-\frac{1}{10}\)
\(B=\frac{7}{30}\)
\(B=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)
\(\Rightarrow B=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(\Rightarrow B=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow B=\frac{1}{3}-\frac{1}{4}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(\Rightarrow B=\frac{1}{12}-\frac{6}{40}\)
\(\Rightarrow B=\frac{-1}{15}\)
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
Mk làm bài này trên cơ sở bài làm của bạn:
\(A=\frac{7}{3.7}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{10}{9.10}\)
\(A=\frac{8-1}{3.4}-\frac{10-1}{4.5}+\frac{12-1}{5.6}-\frac{14-1}{6.7}+\frac{16-1}{7.8}-\frac{18-1}{8.9}+\frac{20-1}{9.10}\)
\(A=\frac{8}{3.4}-\frac{1}{3.4}+\frac{12}{5.6}-\frac{1}{5.6}+\frac{14}{6.7}-\frac{1}{6.7}+\frac{16}{7.8}-\frac{1}{7.8}+\frac{18}{8.9}-\frac{1}{8.9}+\frac{20}{9.10}-\frac{1}{9.10}\)
\(A=\left(\frac{8}{3.4}+\frac{12}{5.6}+\frac{14}{6.7}+\frac{16}{7.8}+\frac{18}{8.9}+\frac{20}{9.10}\right)-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\left(\frac{2.4}{3.4}+\frac{2.6}{5.6}+\frac{2.7}{6.7}+\frac{2.9}{8.9}+\frac{2.10}{9.10}\right)-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(\frac{2}{3}+\frac{2}{5}+\frac{2}{6}+\frac{2}{8}+\frac{2}{9}\right)-\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(A=\left[\left(\frac{2}{3}+\frac{2}{9}\right)+\left(\frac{2}{6}+\frac{2}{8}\right)+\frac{2}{5}\right]-\frac{7}{30}\)
\(A=\left(\frac{8}{9}+\frac{7}{12}+\frac{2}{5}\right)-\frac{7}{30}\)
\(A=\left(\frac{160}{180}+\frac{105}{180}+\frac{72}{180}\right)-\frac{42}{180}\)
\(A=\frac{337}{180}-\frac{42}{180}\)
\(A=\frac{295}{180}=\frac{59}{36}\)
Nguyễn Huy Tú : Cô mình nói đáp án đúng là \(\frac{13}{30}\) còn đáp án của bạn khác với đáp án của cô mình.
bị nhầm xíu sửa lại
\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-...-\frac{1}{2.3}-\frac{1}{1.2}=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-\frac{1}{1}+\frac{1}{2}\)
\(=\frac{2}{9}-\frac{1}{10}-\frac{1}{1}=\frac{20}{90}-\frac{9}{90}-\frac{90}{90}=-\frac{79}{90}\)