\(\dfrac{17}{40}=\dfrac{7}{10}-x+\dfrac{13}{20}\)
\(\Rightarrow\dfrac{17}{40}=\dfrac{28}{40}-x+\dfrac{26}{40}\)
\(\Rightarrow\dfrac{17}{40}=\dfrac{44}{40}-x\)
\(\Rightarrow x=\dfrac{44}{40}-\dfrac{17}{40}\)
\(\Rightarrow x=\dfrac{27}{40}\)
A=\(\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0.9}{\dfrac{7}{91}+0.2-\dfrac{3}{10}}\)
giúp mk nha 
1 tìm x, y biết:
a)\(\dfrac{x+5}{x}\)=\(\dfrac{4}{3}\)
b)\(\dfrac{x-20}{x-10}\)=\(\dfrac{x+40}{x+70}\)
c) 4x=\(\dfrac{1.2.3...31}{4.6.8...64}\)
2 tìm chữ số tận cùng:
A=32017 × 72018 × 132019
-(\(\dfrac{3}{17}\)+\(\dfrac{2}{13}\))-(\(\dfrac{-20}{17}\)+\(\dfrac{11}{13}\)) Tính nhanh từng bước ra nha
Tìm x biết :
\(\dfrac{3}{\left(x+2\right)\cdot\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\cdot\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\cdot\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\cdot\left(x+17\right)}\) Vs x \(\notin\){-2;-5;-17;-10}
TÍNH:
1)\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
2)\(\dfrac{_{-5}}{11}.\dfrac{13}{17}-\dfrac{5}{11}.\dfrac{4}{17}\)
5. \(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\) 6. \(3\dfrac{1}{14}-5\dfrac{1}{3}-\dfrac{4}{7}+\dfrac{11}{21}=x-\dfrac{1}{2}\) 7. \(\dfrac{11}{-40}+\dfrac{4}{5}-\left|\dfrac{3}{4}-1\dfrac{5}{12}\right|=\dfrac{3}{20}-x\)
8) \(\dfrac{17}{-26}.\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}.\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}.\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
5. \(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\) 6. \(3\dfrac{1}{14}-5\dfrac{1}{3}-\dfrac{4}{7}+\dfrac{11}{21}=-\dfrac{1}{2}\) 7. \(\dfrac{11}{-40}+\dfrac{4}{5}-\left|\dfrac{3}{4}-1\dfrac{5}{12}\right|=\dfrac{3}{20}-X\)
So sánh
a, \(\dfrac{2005^{2014}+1}{2005^{2015}+1}\&\dfrac{2005^{2016}+1}{2005^{2017}+1}\)
b, \(\dfrac{19}{10}\&\dfrac{49}{40}\)
c, \(\dfrac{13}{20}\&\dfrac{33}{40}\)
