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\(A=-\frac{5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{12}{7}\\ =-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\\ =-\frac{5}{7}.1+\frac{12}{7}\\ =\frac{12}{7}-\frac{5}{7}\\ =\frac{12-5}{7}\\ =\frac{7}{7}=1\)
Chúc bạn học tốt!
A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
b: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{13}{4}=\dfrac{-5}{7}+\dfrac{13}{4}=\dfrac{-20+91}{28}=\dfrac{71}{28}\)
c: \(=\dfrac{146}{13}-3-\dfrac{68}{13}=6-3=3\)
d: \(=\dfrac{2}{7}\left(\dfrac{21}{4}-\dfrac{13}{4}\right)=\dfrac{4}{7}\)
a: \(=\dfrac{-2}{7}\cdot\dfrac{3}{2}=\dfrac{-3}{7}\)
b: \(=3\cdot\dfrac{7}{12}=\dfrac{7}{4}\)
c: \(=\dfrac{11}{12}\cdot\dfrac{16}{33}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{4}{3}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{4}{5}=\dfrac{4}{15}\)
Lời giải:
\(A=\left\{\frac{k}{2k+1}|k\in\mathbb{N}; 1\leq k\leq 4\right\}\)
Câu 1:
a) = \(\dfrac{-7}{2}\) x \(\dfrac{45}{32}\) = \(\dfrac{-315}{64}\)
b) = \(\dfrac{18}{7}\) : \(\dfrac{-27}{14}\) = \(\dfrac{18}{7}\) x \(\dfrac{14}{-27}\) = \(\dfrac{-4}{3}\)
c) = \(\dfrac{-3}{8}\) x ( \(\dfrac{5}{11}\) + \(\dfrac{6}{11}\) + 2 ) = \(\dfrac{-3}{8}\) x 3 = \(\dfrac{-9}{8}\)
Câu 2:
\(\dfrac{-3}{5}\) . x + \(\dfrac{7}{6}\) = \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{5}{4}\) - \(\dfrac{7}{6}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{1}{12}\)
\(\Leftrightarrow\) x = \(\dfrac{1}{12}\) : \(\dfrac{-3}{5}\)
\(\Leftrightarrow\) x = \(\dfrac{-5}{36}\)
\(\dfrac{2}{7}\cdot\dfrac{2}{11}+\dfrac{5}{7}\cdot\dfrac{12}{11}-\dfrac{5}{7}\cdot\dfrac{7}{11}\)
\(=\dfrac{4}{77}+\dfrac{5}{7}\left(\dfrac{12}{11}-\dfrac{7}{11}\right)\)
\(=\dfrac{4}{77}+\dfrac{5}{7}\cdot\dfrac{5}{11}=\dfrac{4}{77}+\dfrac{25}{77}=\dfrac{29}{77}\)
p/s: toán 6
ai có thể dạy mk cách viết p/s đc ko???