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Ta có :\(A=\left(5+10+15+...+1000\right).\left\{\frac{2}{5}:0,5+2:\left(-0,4\right)\right\}:\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1000}\right)\)
\(\Leftrightarrow A=\left(5+10+15+...+1000\right).\left\{\frac{2}{5}:\frac{1}{2}+2:\left(-\frac{2}{5}\right)\right\}:\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1000}\right)\)\(\Leftrightarrow A=\left(5+10+15+...+1000\right).\left\{\frac{2}{5}.2+2.\left(-\frac{5}{2}\right)\right\}:\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1000}\right)\)\(\Leftrightarrow A=\left(5+10+...+1000\right).\left\{2.\left(\frac{2}{5}-\frac{5}{2}\right)\right\}.\left(5+10+...+1000\right)\)
\(\Leftrightarrow A=\left(5+10+...+1000\right).\left(5+10+...+1000\right).-\frac{21}{10}\)
Ta có : Số số hạng của dãy số : \(5+10+...+1000\) là :
\(\left(1000-5\right):5+1=200\)
\(\Rightarrow\) Tổng của dãy số : \(5+10+...+1000\) là :
\(\frac{\left(5+1000\right).200}{2}=100500\)
\(\Rightarrow A=100500.100500.\left(-\frac{21}{10}\right)\)
\(\Rightarrow A=100500^2.\left(-\frac{21}{10}\right)\)
\(\Rightarrow A=\frac{100500^2.\left(-21\right)}{10}\)
Vậy :\(A=\frac{100500^2.\left(-21\right)}{10}\)
P/s: Số to quá nên mình đề dưới dạng phân số, không tính ra kết quả cụ thể.
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
a: \(\left(0.5\right)^3\cdot2^3=1\)
b: \(\left(0.25\right)^2\cdot16=1\)
c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)
a. \(x-\dfrac{2}{35}=\dfrac{-2}{25}\)
<=> \(x=\dfrac{-2}{25}+\dfrac{2}{35}\)
<=> x = \(-\dfrac{4}{175}\)
b. \(\dfrac{11}{12}-\left(x+\dfrac{2}{5}\right)=\dfrac{2}{3}\)
<=> \(\dfrac{11}{12}-x-\dfrac{2}{5}=\dfrac{2}{3}\)
<=> \(-x=\dfrac{2}{3}-\dfrac{11}{12}+\dfrac{2}{5}\)
<=> \(-x=\dfrac{3}{20}\)
<=> \(x=\dfrac{-3}{20}\)
c. \(\dfrac{2}{5}+\dfrac{1}{4}x=\dfrac{-3}{10}\)
<=> \(\dfrac{1}{4}x=\dfrac{-3}{10}-\dfrac{2}{5}\)
<=> \(\dfrac{1}{4}x=\dfrac{-7}{10}\)
<=> \(x=\dfrac{-14}{5}\)
a) \(x-\dfrac{2}{35}=-\dfrac{2}{25}\)
\(\Rightarrow x=-\dfrac{2}{25}+\dfrac{2}{35}\)
\(\Rightarrow x=-\dfrac{4}{175}\)
b) \(\dfrac{11}{12}-\left(x+\dfrac{2}{5}\right)=\dfrac{2}{3}\)
\(\Rightarrow x+\dfrac{2}{5}=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\Rightarrow x+\dfrac{2}{5}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(\Rightarrow x=-\dfrac{3}{20}\)
c) \(\dfrac{2}{5}+\dfrac{1}{4}x=-\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{7}{10}\)
\(\Rightarrow x=-\dfrac{7}{10}:\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{14}{5}\)