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\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)
x=0.78
a ) ( - 14 ) + 12 + ( - 12 ) . ( 23 - 7 )
= - 2 + ( - 12 ) . ( 8 - 7 )
= - 2 + ( - 12 ) . 1
= - 2 + ( - 12 )
= - 14
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
Bài 1:
\(=\dfrac{-3-39}{32}+\dfrac{-6-11}{17}+\dfrac{-1}{6}=-\dfrac{21}{16}+\dfrac{-1}{6}-1=-\dfrac{119}{48}\)
Bài 2:
\(\Leftrightarrow x:5=-\dfrac{13}{20}\)
hay x=-13/4
`Answer:`
\(12x-144=0\)
\(\Leftrightarrow12x=144\)
\(\Leftrightarrow x=144:12\)
\(\Leftrightarrow x=12\)
\(5x-32:18=13\)
\(\Leftrightarrow5x-\frac{16}{9}=13\)
\(\Leftrightarrow5x=13+\frac{16}{9}\)
\(\Leftrightarrow5x=\frac{133}{9}\)
\(\Leftrightarrow x=\frac{133}{9}:5\)
\(\Leftrightarrow x=\frac{133}{45}\)
\(3x+6=15\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\)
