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\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{39}=\frac{8}{9}.\)
\(\Rightarrow x.\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{39}\right)=\frac{8}{9}\)
\(\Rightarrow x.\frac{23}{52}=\frac{8}{9}\)
\(\Rightarrow x=\frac{426}{207}\)
Study well
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{39}=\frac{8}{9}\)
<=> x\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{x}{39}\right)=\frac{8}{9}\)
<=> x = \(\frac{8}{9}:\frac{23}{52}\)
<=> x = \(\frac{416}{207}\)
(1-1/2).(1.1/3).(1.1-4)....(1-1/2016)
=1/2.2/3.3/4.4/5....2015/2016
=1.2.3....2015/2.3.4....2016
=1/2016
\(=\frac{4}{8}+\frac{6}{8}-\frac{1}{8}\)
\(=\frac{9}{8}\)
học tốt
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(=\frac{4}{8}+\frac{2}{8}+\frac{1}{8}\)
\(=\frac{7}{8}\)
Study well
a,Chiều dài là:
60.4/3=80(m)
Diện tích khu vườn là:
80.60=4800(m2)
b,Diện tích trông cây ăn quả là:
4800.7/12=2800(m2)
Diện tích còn lại là:
4800-2800=2000(m2)
Diện tích đào ao là:
2000.30%=600(m2)
Đáp số : a,4800m2 b,600m2
\(\left|2x-1\right|-5=\frac{3}{4}\)
\(=>\left|2x-1\right|=\frac{23}{4}\)
\(=>2x-1=\pm\frac{23}{4}\)
\(\hept{\begin{cases}2x-1=\frac{23}{4}=>x=\frac{27}{8}\\2x-1=-\frac{23}{4}=>x=-\frac{19}{8}\end{cases}}\)
\(\frac{2}{3}\cdot x+\frac{3}{7}=-\frac{1}{2}\)
\(=>\frac{2}{3}\cdot x=-\frac{1}{14}\)
\(=>x=-\frac{3}{28}\)
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}=\frac{8}{9}\)
\(\Leftrightarrow\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}=\frac{8}{18}=\frac{4}{9}\)
\(\Leftrightarrow\frac{x}{3.4}+\frac{x}{4.5}+\frac{x}{5.6}+\frac{x}{6.7}+\frac{x}{7.8}+\frac{x}{8.9}=\frac{4}{9}\)
\(\Leftrightarrow\frac{x}{3}-\frac{x}{4}+\frac{x}{4}-\frac{x}{5}+\frac{x}{5}-\frac{x}{6}+...+\frac{x}{8}-\frac{x}{9}=\frac{4}{9}\)
\(\Leftrightarrow\frac{x}{3}-\frac{x}{9}=\frac{4}{9}\Leftrightarrow\frac{2x}{9}=\frac{4}{9}\Leftrightarrow x=2\)
\(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(1+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\left(đpcm\right)\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}\)
\(B=\frac{1-\frac{1}{3^{2005}}}{2}\)
\(B=\frac{1}{2}-\frac{1}{\frac{3^{2005}}{2}}\)
Vi \(\frac{1}{2}-\frac{1}{\frac{3^{2005}}{2}}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\left(dpcm\right)\)
Ta có : \(\widehat{O_1}\)và \(\widehat{O_3}\) ( đối đỉnh )
\(\widehat{O_1}\) và \(\widehat{O_3}=130^o\left(gt\right)\)
\(\Rightarrow\widehat{O_1}=\widehat{O_3}=\frac{130^o}{2}=65^o\)
\(\widehat{O_2}=\widehat{O_4}=180^o-65^o=115^o\)
Chúc bạn học tốt !!!