Tìm B biết:
B= \(\dfrac{\text{6^10 x 3 - 4^4 x 9^5 x 84 }}{\text{2^13 x 3^10 + 12^5 x 4 x 3^6}}\) .
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c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
1. Tính
\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)
\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)
2. Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)
\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)
\(=\dfrac{29}{35}+\dfrac{3}{4}\)
\(=\dfrac{116}{140}+\dfrac{105}{140}\)
\(=\dfrac{221}{140}\)
\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)
\(=\dfrac{9}{7}-\dfrac{55}{77}\)
\(=\dfrac{99}{77}-\dfrac{55}{77}\)
\(=\dfrac{44}{77}=\dfrac{4}{7}\)
\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)
\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(=\dfrac{27}{35}\)
\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}\times\dfrac{11}{3}\)
\(=\dfrac{154}{135}\)
\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)
\(=\dfrac{41}{21}-\dfrac{1}{4}\)
\(=\dfrac{164}{84}-\dfrac{21}{84}\)
\(=\dfrac{143}{84}\)
\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)
\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)
\(=\dfrac{20}{27}\times\dfrac{2}{11}\)
\(=\dfrac{40}{297}\)
\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\dfrac{32}{10}:\dfrac{2}{5}\)
\(=\dfrac{16}{5}\times\dfrac{5}{2}\)
\(=\dfrac{80}{10}=8\)
a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)
b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)
c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)
d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
\(B=\dfrac{6^{10}\cdot3-4^4\cdot9^5\cdot84}{2^{13}\cdot3^{10}+12^5\cdot4\cdot3^6}\)
\(B=\dfrac{3^{10}\cdot2^{10}\cdot3-\left(2^2\right)^4\cdot\left(3^2\right)^5\cdot2^2\cdot21}{2^{13}\cdot3^{10}+3^5\cdot4^5\cdot4\cdot3^6}\)
\(B=\dfrac{3^{11}\cdot2^{10}-2^8\cdot3^{10}\cdot2^2\cdot3\cdot7}{2^{13}\cdot3^{10}+3^{11}\cdot2^{12}}\)
\(B=\dfrac{3^{11}\cdot2^{10}-2^{10}\cdot3^{11}\cdot7}{2^{13}\cdot3^{10}+3^{11}\cdot2^{12}}\)
\(B=\dfrac{3^{11}\cdot2^{10}\cdot\left(1-7\right)}{2^{12}\cdot3^{10}\cdot\left(2+3\right)}\)
\(B=\dfrac{2^{10}\cdot3\cdot-6}{2^{12}\cdot5}\)
\(B=\dfrac{3\cdot-6}{2^2\cdot5}\)
\(B=\dfrac{-18}{20}\)
\(B=-\dfrac{9}{10}\)