a) \(\frac{-1}{39}+\frac{-1}{52}\) b)\(\frac{-6}{9}+\frac{-12}{16}\) C)\(\frac{-2}{5}-\frac{-3}{11}\) D) \(\frac{-34}{37}\times\frac{74}{-85}\)
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HQ
Hà Quang Minh
Giáo viên
2 tháng 11 2023
a) $\frac{1}{6} \times \frac{2}{3} = \frac{{1 \times 2}}{{6 \times 3}} = \frac{2}{{18}} = \frac{1}{9}$
b) $\frac{6}{5} \times \frac{3}{8} = \frac{{6 \times 3}}{{5 \times 8}} = \frac{{18}}{{40}} = \frac{9}{{20}}$
c) $\frac{4}{3} \times \frac{8}{9} = \frac{{4 \times 8}}{{3 \times 9}} = \frac{{32}}{{27}}$
d) $\frac{5}{{12}} \times \frac{{12}}{5} = \frac{{5 \times 12}}{{12 \times 5}} = \frac{{60}}{{60}} = 1$
29 tháng 6 2022
a: \(=-9+\left\{-52:9\right\}=-9+\dfrac{-52}{9}=-\dfrac{133}{9}\)
b: \(=\dfrac{17}{7}+\left(\dfrac{-76}{63}\right):15\)
\(=\dfrac{17}{7}-\dfrac{76}{63}\cdot\dfrac{1}{15}=\dfrac{317}{135}\)
e: \(=\dfrac{-5}{13}\cdot\dfrac{7}{3}-\dfrac{2}{7}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)
\(=\dfrac{5}{13}\left(-\dfrac{7}{3}+\dfrac{1}{7}\right)-\dfrac{2}{7}\cdot\dfrac{8}{13}\)
\(=\dfrac{5}{13}\cdot\dfrac{-46}{21}-\dfrac{16}{91}=\dfrac{-278}{273}\)
25 tháng 1 2017
Bài 1:
\(B=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left(0,625-0,5+\frac{5}{11}+\frac{5}{12}\right)}+\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left[5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)\right]}+\frac{3}{5}\)
\(=\frac{-3}{5}+\frac{3}{5}\)
\(=0\)
Bài 2:
b) Giải:
Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{c^6}{d^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (1)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{b+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^6=\left(\frac{a+c}{b+d}\right)^6=\frac{a^6}{b^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\left(đpcm\right)\)
18 tháng 3 2020
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
18 tháng 3 2020
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
a, \(\frac{-1}{39}+\frac{-1}{52}=\frac{-61}{468}\)
b, \(-\frac{9}{6}+-\frac{12}{16}=-\frac{9}{4}\)
\(c,-\frac{2}{5}--\frac{3}{11}=-\frac{7}{55}\)\
\(-\frac{34}{37}\cdot\frac{74}{-58}=\frac{34}{29}\)