Tính A/B biết:
A=1/2+1/3+1/4+...+1/5 ; B=1/199+2/198+1/197+...+1.98/2+199/1
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1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)
=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)
=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)
=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)
=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)
b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)
\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)
\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)
\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)
a, \(\dfrac{1}{7}\) x ( \(\dfrac{4}{3}\))2 - \(\dfrac{1}{7}\) : \(\dfrac{9}{11}\)
= \(\dfrac{1}{7}\) x \(\dfrac{16}{9}\) - \(\dfrac{1}{7}\) x \(\dfrac{11}{9}\)
= \(\dfrac{1}{7}\) x ( \(\dfrac{16}{9}-\dfrac{11}{9}\))
= \(\dfrac{1}{7}\) x \(\dfrac{5}{9}\)
= \(\dfrac{5}{63}\)
b, 2x + \(\dfrac{1}{4}\) = \(\dfrac{3}{5}\)
2x = \(\dfrac{3}{5}-\dfrac{1}{4}\)
2x = \(\dfrac{7}{20}\)
x = \(\dfrac{7}{20}:2\)
x = \(\dfrac{7}{40}\)
a) Đặt A=1/2 + 1/4 + 1/8 +...+ 1/256 + 1/512
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
b)\(\frac{a}{b}+\frac{4}{6}+\frac{2}{10}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b}+\frac{13}{15}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b}=\frac{19}{30}\)
\(\frac{4}{5}:\frac{a}{b}-\frac{6}{5}=\frac{3}{10}\)
\(\Rightarrow\frac{4}{5}:\frac{a}{b}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b}=\frac{8}{15}\)
các pạn giúp mk với mk đang cần gấp!
Em nên xem lại câu hỏi vì nó khá khó hiểu A và B được xác định theo quy luật nào