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1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(=\frac{\frac{1}{7}\left(\frac{1}{2}-\sqrt{2}+\frac{3\sqrt{2}}{5}\right).\frac{-4}{15}}{\frac{1}{5}\left(\frac{1}{2}+\frac{3\sqrt{2}}{5}-\sqrt{2}\right).\frac{5}{7}}\)
\(=\frac{\frac{1}{7}.\frac{-4}{15}}{\frac{1}{5}.\frac{5}{7}}=\frac{\frac{-4}{105}}{\frac{1}{7}}=\frac{-4}{15}\)
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3.\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{10}.\frac{5}{7}+\frac{3.\sqrt{2}}{25}.\frac{5}{7}-\frac{\sqrt{2}}{5}.\frac{5}{7}}\)
\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{14}+\frac{3.\sqrt{2}}{35}-\frac{\sqrt{2}}{7}}\)
\(=\frac{-4}{15}\)
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