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Đăt \(A=\dfrac{40}{31.39}+\dfrac{35}{39.46}+\dfrac{30}{46.52}+\dfrac{25}{52.57}+\dfrac{20}{57.61}\)
\(\Leftrightarrow A=5.\left(\dfrac{8}{31.39}+\dfrac{7}{39.46}+\dfrac{6}{46.52}+\dfrac{5}{52.57}+\dfrac{4}{57.61}\right)\)
\(\Leftrightarrow A=5.\left(\dfrac{1}{31}-\dfrac{1}{39}+\dfrac{1}{39}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{52}+\dfrac{1}{52}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{61}\right)\)\(\Leftrightarrow A=5.\left(\dfrac{1}{31}-\dfrac{1}{61}\right)\)
\(\Leftrightarrow A=5.\dfrac{30}{31.61}\)
\(\Leftrightarrow A=\dfrac{150}{1891}\)
Lời giải:
\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)
\(\Rightarrow A=\frac{20}{21}\)
\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{30}{31}\)
\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)
Do đó $A>B$
A.2=4/1.5+6/5.11+...+12/29.41
A.2=1-1/5+1/5-1/11+...+1/29-1/41
A.2=1-1/41
A.2=40/41
A=20/41
B.3=3/1.4+6/4.10+...+12/29.31
B.3=1-1/4+1/4-1/10+...+1/29-1/31
B.3=1-1/31
B.3=30/31
B=10/31
Vì 20/41.10/31 nên A>B
\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(\Rightarrow2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)
\(\Rightarrow2A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)
\(\Rightarrow2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)
\(\Rightarrow A=\dfrac{40}{41}:2=\dfrac{20}{41}\)(1)
\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
\(\Rightarrow3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)
\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\)
\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{31}=\dfrac{30}{31}\)
\(\Rightarrow B=\dfrac{30}{31}:3=\dfrac{10}{31}\)
\(\Rightarrow B=\dfrac{2}{2}.\dfrac{10}{31}=\dfrac{20}{62}\)
+)Ta có:\(\dfrac{20}{62}< \dfrac{20}{41}\Rightarrow B< A\)
Hay A>B(ĐPCM)
Chúc bn học tốt
\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)
a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)
b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)
\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)
\(=\dfrac{20}{720}=\dfrac{1}{36}\)
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)
\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)
\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)
\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)
\(=-\dfrac{69}{45}\)
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)