Chứng minh A > B, biết A = \(\dfrac{2}{5.7}+\dfrac{5}{7.12}+\dfrac{7}{12.19}+\dfrac{9}{19.28}+\dfrac{11}{28.39}+\dfrac{1}{39.40}\)
B = \(\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}\)
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Ta có : +) A= 1/5 -1/7 +1/7 -1/12 +1/12 - 1/19 +1/19 - 1/28 +1/28 - 1/39 +1/30.40 ⇔ A=1/5 -1/39 +1/30.40
+) B= 2.(1/5.8 +1/8.11 +1/11.14 +1/14.17 + 1/17.20 )
⇔B=2. 1/3.(1/5 - 1/8 +1/8 - 1/11 +1/11- 1/14 +1/14 -1/17 +1/17 -1/20 )
⇔B=2/3.( 1/5-1/20 ) Ta luôn có :B luôn <1/5 - 1/20
Mà 1/5 -1/20 <1/5 -1/39 +1/30.40 =A
⇒ A>B (dpcm) Tích mình với nha bn .
\(A=\frac{2}{5.7}+\frac{5}{7.12}+\frac{7}{12.19}+\frac{9}{19.28}+\frac{11}{28.39}+\frac{1}{39.40}\)
\(=\frac{7-5}{5.7}+\frac{12-7}{7.12}+\frac{19-12}{12.19}+\frac{28-19}{19.28}+\frac{39-28}{28.39}+\frac{40-39}{39.40}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{19}+\frac{1}{19}-\frac{1}{28}+\frac{1}{28}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(=\frac{1}{5}-\frac{1}{40}=\frac{7}{40}\)
\(B=\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}\)
\(=\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}+\frac{2}{340}\)
\(=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}+\frac{2}{17.20}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{2}{3}\left(\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{1}{10}\)
\(\frac{A}{B}=\frac{\frac{7}{40}}{\frac{1}{10}}=\frac{7}{4}\)
Câu C giải rồi
\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)
\(\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
=\(\dfrac{2}{10}+\dfrac{2}{40}+\dfrac{2}{88}+\dfrac{2}{154}+\dfrac{2}{238}+\dfrac{2}{340}+\dfrac{2}{460}+\dfrac{2}{598}\)
=\(\dfrac{1}{3}.2\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
=\(\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
=\(\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)\)
=\(\dfrac{2}{3}.\dfrac{6}{13}\)
=\(\dfrac{4}{13}\)
\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)
\(x=\dfrac{7}{35}+\dfrac{-15}{35}\)
\(x=-\dfrac{8}{35}\)
\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{2}\)
\(x=\dfrac{4}{7}:\dfrac{3}{2}\)
\(x=\dfrac{4}{7}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{21}\)
\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)
\(x+\dfrac{3}{4}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}-\dfrac{3}{4}\)
\(x=-\dfrac{23}{12}\)
\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{13}{18}\)
\(x=\dfrac{-5}{9}-\dfrac{13}{18}\)
\(x=\dfrac{-10}{18}-\dfrac{13}{18}\)
\(x=-\dfrac{23}{18}\)
a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)
b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)
c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)
ta tách 2/5x7 = 2/5-2/7 tách những cái kia tương tự góp vào rồi tính
bn lm sai ùi