\(\dfrac{4}{6} - \dfrac{2}{4} =.......\)
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a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}\right)\)
\(=0+0+0+0-\dfrac{257}{210}\)
\(=\dfrac{257}{210}\)
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
\(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
\(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{49}{30}\)
\(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\)
\(\dfrac{8}{5}x\dfrac{5}{8}=\dfrac{1}{1}=1\)
\(\dfrac{6}{7}x\dfrac{4}{7}=\dfrac{24}{49}\)
\(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}x\dfrac{5}{4}=\dfrac{1}{1}=1\)
\(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}x\dfrac{5}{5}=\dfrac{1}{1}=1\)
1) \(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{1+2}{3}=\dfrac{3}{3}=1\)
2) \(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{24+25}{30}=\dfrac{49}{30}\)
3) \(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{4-3}{5}=\dfrac{1}{5}\)
4) \(\dfrac{9}{8}-\dfrac{4}{2}=\dfrac{9}{8}-2=\dfrac{9}{8}-\dfrac{16}{8}=-\dfrac{7}{8}\)
5) \(\dfrac{8}{5}\times\dfrac{5}{8}=\dfrac{8\times5}{5\times8}=\dfrac{40}{40}=1\)
6) \(\dfrac{6}{7}\times\dfrac{4}{7}=\dfrac{6\times4}{7}=\dfrac{24}{7}\)
7) \(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{5}{4}=\dfrac{4\times5}{5\times4}=\dfrac{20}{20}=1\)
8) \(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}\times\dfrac{5}{5}=\dfrac{5\times5}{5\times5}=\dfrac{25}{25}=1\)
a: =>-10<3x<-7
mà x là số nguyên
nên 3x=-9
hay x=-3
b: =>-3<2x<-2
mà x là số nguyên
nên \(x\in\varnothing\)
a: \(=\dfrac{-8}{9}-\dfrac{6}{5}+\dfrac{8}{9}=-\dfrac{6}{5}\)
c: \(=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)
k/
\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)
\(\Leftrightarrow96-4x+8=3x\)
\(\Leftrightarrow96-4x+8-3x=0\)
\(\Leftrightarrow104-7x=0\)
\(\Leftrightarrow7x=104\)
\(\Leftrightarrow x=104:7\)
\(\Leftrightarrow x=\dfrac{104}{7}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)
m/
\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow9x+6-3x-1-12x-10=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)
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