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a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
\(\dfrac{7}{6}\) + \(\dfrac{5}{12}\) - \(\dfrac{1}{18}\) - 1
= \(\dfrac{42}{36}\) + \(\dfrac{15}{36}\) - \(\dfrac{2}{36}\) - \(\dfrac{36}{36}\)
= \(\dfrac{19}{36}\)
\(\dfrac{13}{6}\) + \(\dfrac{5}{18}+3-\dfrac{7}{12}\)
= \(\dfrac{78}{36}+\dfrac{10}{36}+\dfrac{108}{36}-\dfrac{21}{36}\)
= \(\dfrac{175}{36}\)
3 + \(\dfrac{11}{4}-\dfrac{1}{12}-\dfrac{3}{16}\)
= \(\dfrac{144}{48}+\dfrac{132}{48}-\dfrac{4}{48}-\dfrac{9}{48}\)
= \(\dfrac{263}{48}\)
\(\dfrac{2}{5}\) x \(\dfrac{7}{4}\) - \(\dfrac{2}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{2}{5}\) x ( \(\dfrac{7}{4}\) - \(\dfrac{3}{4}\))
= \(\dfrac{2}{5}\) x \(\dfrac{4}{4}\)
= \(\dfrac{2}{5}\)
\(\left(390:x-90:x\right)=3\)
\(\Rightarrow\left(390-90\right):x=3\)
\(\Rightarrow300:x=3\)
\(\Rightarrow x=300:3\)
\(\Rightarrow100\)
\(8765\div90=\dfrac{1753}{18}\)\(=97,4\)
\(1325\times65=86125\)
\(6543\div89=73,52\)
Ta có: 1/2 + 1/6 + 1/12 + ... + 1/90
= 1/1.2 + 1/2.3 +1/3.4 +...+ 1/9.10
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... +1/9 - 1/10
= 1 - 1/10
= 9/10
Nếu đúng thì cho mk biết nha
\(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
em thưa cô đề này là em dc cô giáo học thêm cho và cô chưa dậy nên em nghĩ là đúng đề rồi ạ
100 = 600 nhé
600 nhóe:)