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\(=\left(\dfrac{7}{18}+\dfrac{3}{34}\right):\left[\dfrac{14}{4}\left(\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{18}+\dfrac{1}{13}-\dfrac{1}{17}\right)\right]\)
\(=\dfrac{73}{153}:\left[\dfrac{14}{4}\cdot\dfrac{73}{1071}\right]\)
\(=\dfrac{73}{153}:\dfrac{73}{306}=2\)
a) Ta có: \(A=\left(\dfrac{63}{9\cdot18}+\dfrac{21}{14\cdot17}\right):\left(\dfrac{14}{9\cdot13}+\dfrac{14}{14\cdot18}+\dfrac{14}{3\cdot17}\right)\)
\(=\left(7\cdot\dfrac{9}{9\cdot18}+7\cdot\dfrac{3}{14\cdot17}\right):\left(14\left(\dfrac{1}{9\cdot13}+\dfrac{1}{14\cdot18}+\dfrac{1}{3\cdot17}\right)\right)\)
\(=\dfrac{7\left(\dfrac{1}{9}-\dfrac{1}{18}+\dfrac{1}{14}-\dfrac{1}{17}\right)}{14\cdot\dfrac{1}{4}\cdot\left(\dfrac{4}{9\cdot13}+\dfrac{4}{3\cdot17}+\dfrac{4}{14\cdot18}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{18}+\dfrac{1}{14}-\dfrac{1}{17}}{\dfrac{1}{2}\left(\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{3}-\dfrac{1}{17}+\dfrac{1}{14}-\dfrac{1}{18}\right)}\)
\(=\dfrac{\dfrac{73}{1071}}{\dfrac{1}{2}\cdot\dfrac{4519}{13923}}=\dfrac{1898}{4519}\)
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
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