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Đề Này Không tính nhanh được đâu bạn
Minh tính Ra kết quả là :
\(\frac{2257}{378}\)
a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)
d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right).\)
\(=5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+\frac{73}{35}-\frac{5}{6}-8-\frac{2}{7}+\frac{1}{18}\)
\(=\left(5-2-8\right)+\left(\frac{1}{5}+\frac{73}{35}-\frac{2}{7}\right)+\left(-\frac{2}{9}-\frac{5}{6}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\left(\frac{7}{35}+\frac{73}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\frac{70}{35}+\frac{-18}{18}+\frac{1}{23}\)
\(=-5+2+\left(-1\right)+\frac{1}{23}\)
\(=-4+\frac{1}{23}=-\frac{91}{23}\)
Tính nhanh nếu có thể:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)\)
\(=\)\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-\frac{73}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)\)
\(=\)\(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+\frac{73}{35}-\frac{5}{6}-8-\frac{2}{7}+\frac{1}{18}\)
\(=\)\(\left(5-2-8\right)+\left(\frac{1}{5}+\frac{73}{35}-\frac{2}{7}\right)+\left(\frac{2}{9}-\frac{5}{6}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\)\(\left(-5\right)+\left(\frac{7}{35}+\frac{73}{35}-\frac{10}{35}\right)+\left(\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\)\(\left(-5\right)+2+-\frac{5}{9}+\frac{1}{23}\)
\(=\)\(\left(-3\right)+-\frac{5}{9}+\frac{1}{23}\)
\(=\)\(-\frac{727}{207}\)
