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=3/2x( 1/3-1/5+1/5-1/7-1/7+....+ 1/97-1/99)
=3/2x( 1/3-1/99)
=3/2x 32/99
= 16/33
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a \(\frac{4}{18}+\frac{5}{13}+\frac{7}{9}+\frac{21}{13}\)
= \(\left(\frac{4}{18}+\frac{7}{9}\right)+\left(\frac{5}{13}+\frac{21}{13}\right)\)
= \(1+2\)
= \(3\)
b \(\frac{4}{3}-\frac{2}{7}-\frac{5}{7}+\frac{2}{3}\)
= \(\left(\frac{4}{3}-\frac{2}{3}\right)-\left(\frac{2}{7}+\frac{5}{7}\right)\)
= \(\frac{2}{3}-1\)
= \(-\frac{1}{3}\)
a) 2/9 +1/5 +7/9+4/5
=( 2/9+7/9)+(1/5+4/5)
=1+1=2
b) 1/12+3/16+5/12+5/16
=(1/12 +5/12)+(3/16+5/16)
=1/2 +1/2=1
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)
Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)
Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)
Bài làm :
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}\)
\(A>\frac{2}{5}\left(1\right)\)
Ta cũng có :
\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}\)
\(A< \frac{8}{9}\left(2\right)\)
\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)
=> Điều phải chứng minh
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(=14:\left(\frac{14}{3}-\frac{14}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
\(=14:\frac{28}{9}+14:\frac{14}{9}\)
\(=\frac{28}{9}+\frac{14}{9}\)
\(=\frac{42}{9}\)
\(14:\left(\frac{14}{3}-\frac{14}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
\(=14:\left(\frac{42}{9}-\frac{14}{9}\right)+14:\left(\frac{6}{9}+\frac{8}{9}\right)\)
\(=14:\frac{28}{9}+14:\frac{14}{9}\)
\(=14.\frac{9}{28}+14.\frac{9}{14}\)
\(=14.\left(\frac{9}{28}+\frac{9}{14}\right)\)
\(=14.\left(\frac{9}{28}+\frac{18}{28}\right)\)
\(=14.\frac{27}{28}\)
\(=\frac{378}{28}\)
\(=\frac{27}{2}\)
= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua