\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+........+\dfrac{209}{210}\)
Giúp mk bài này nha ,mk cần gấp nhé
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`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`
`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`
`=sqrt{10}-(2+sqrt{10})`
`=-2`
`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`
`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`
`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`
`=(5sqrt5-5sqrt3-4)/2`
\(\dfrac{1}{2022}\) \(\times\) \(\dfrac{2}{5}\) + \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{7}{5}\) - \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{8}{10}\)
= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{2}{5}\) + \(\dfrac{7}{5}\) - \(\dfrac{8}{10}\))
= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{9}{5}\) - \(\dfrac{4}{5}\))
= \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{5}{5}\)
= \(\dfrac{1}{2022}\times1\)
= \(\dfrac{1}{2022}\)
\(\dfrac{x}{2}-\dfrac{2}{9}=\dfrac{3}{2}-\left(\dfrac{-5}{4}\right)\\ \Rightarrow\dfrac{x}{2}-\dfrac{2}{9}=\dfrac{11}{4}\\ \Rightarrow\dfrac{x}{2}=\dfrac{11}{4}+\dfrac{2}{9}\\ \Rightarrow\dfrac{x}{2}=\dfrac{107}{36}\)
\(\Rightarrow x=\dfrac{107}{36}\times2\\ \Rightarrow x=\dfrac{107}{18}\)
\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)
\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)
\(\Leftrightarrow x\cdot18=252\)
hay x=14
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)
\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)
\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)
\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)
\(=-\dfrac{69}{45}\)