\(\frac{1+\left(1+2\right)+\left(1+2+3\right)\left(1+2+3+4\right)+...+\left(1+2+3+...+10\right)}{1.99+2.98+...+99.1}\)
Những câu hỏi liên quan
Tính nhanh: \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+99.1}\)
Ta có:Xét tử số
TS có 99 tổng,1 có mặt trong 99 tổng,2 có mặt trong 98 tổng,3 có mặt trong 97 tổng,...,99 có mặt trong 1 tổng
Vì thế ta được:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+1.99}\)
\(=\frac{1.99+2.98+3.97+...+99.1}{1.99+2.98+3.97+...+99.1}=1\)
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Cho \(B=\frac{2.\left(1.99+2.98.+3.97+.....+97.3+98.2+99.1\right)}{\left(1^2+2^2+3^2+.....+97^2+98^2+99^2\right)}\)
So sánh B với 1.
Cho B=\(\frac{2.\left(1.99+2.98+3.97+.....+97.3+98.2+99.1\right)}{\left(1^2+2^2+3^2+.....+97^2+98^2+99^2\right)}\)
So sánh B với 1.
giải cả bài ra cho mình
Cho \(B=\frac{2.\left(1.99+2.98+3.97+...+97.3+98.2+99.1\right)}{1^2+2^2+3^2+...+97^2+98^2+99^2}\)
So sánh B với 1
So sánh biểu thức sau với 1 \(A=\frac{2.\left(1.99+2.98+3.97+4.96+...+97.3+98.2+99.1\right)}{1^2+2^2+3^2+.....+97^2+98^2+99^2}\)
\(Cho:\) \(B=\frac{2\left(1.99+2.98+3.97+...+97.3+98.2+99.1\right)}{\left(1^2+2^2+3^2+...+97^2+98^2+99^2\right)}\)
\(So\)\(sánh\)\(B\)\(với\)\(1\).
left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
Đọc tiếp
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
tính
a) \(3^{-2}.\left[\left(\frac{2}{3}\right)^{-4}\right].\left[\left(-1\frac{1}{2}\right)^{-3}\right]\)
b) \(\left[\left(0.02\right)^{-3}\right].10^{-4}.\left(\frac{4}{5}\right)^{-2}\)
c) \(\left[2^{-2}-\frac{3}{4}^{-4}.\left(\frac{-1}{2}^2\right)\right]:\left(10^{-1}+1\right)\)
các bn ơi câu b là 0,02 chứ không phải 0.02 nhé
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Thực hiện phép tính :
a, A =\(\left(1:\frac{5^2}{10^2}\right).\left(1\frac{1}{1}\right)^2+25.\left[1:\left(\frac{4}{3}\right)^2:\left(\frac{5}{4}\right)^3\right]:\left(1:\frac{-8}{27}\right)\)
b, B =\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{100^2}\right)\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
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