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\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-x=3\sqrt{3}\\\dfrac{2}{3}-x=-3\sqrt{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2-9\sqrt{3}}{3}\\x=\dfrac{2+9\sqrt{3}}{3}\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`-9/34*17/4`
`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)
`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)
`=`\(-\dfrac{9}{8}\)
`b)`
\(\dfrac{17}{15}\div\dfrac{4}{3}\)
`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)
`=`\(\dfrac{17}{20}\)
`c)`
\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)
`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)
a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)
\(=\dfrac{-9\cdot17}{34\cdot4}\)
\(=-\dfrac{153}{136}\)
\(=\dfrac{9}{8}\)
b) \(\dfrac{17}{15}:\dfrac{4}{3}\)
\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)
\(=\dfrac{17\cdot3}{15\cdot4}\)
\(=\dfrac{51}{60}=\dfrac{17}{20}\)
c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)
\(=\dfrac{21}{5}:-\dfrac{14}{5}\)
\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)
\(=\dfrac{21\cdot-5}{5\cdot14}\)
\(=-\dfrac{105}{70}=\dfrac{3}{2}\)
\(\dfrac{x}{3}-2=\dfrac{1}{15}\)
=>\(\dfrac{x}{3}=2+\dfrac{1}{15}=\dfrac{31}{15}\)
=>\(x=\dfrac{31}{15}\cdot3=\dfrac{31}{5}\)
=>(2x-1)^2=24^2
=>2x-1=24 hoặc 2x-1=-24
=>x=-23/2 hoặc x=25/2
Đặt : \(\dfrac{x}{5}=\dfrac{y}{3}=k\)
`=>x=5k,y=3k`
Ta có : \(x^2-y^2=4=>\left(5k\right)^2-\left(3k\right)^2=4\\ =>25k^2-9k^2=4\\ =>16k^2=4\\ =>k^2=\dfrac{1}{4}\\ =>k=\pm\dfrac{1}{2}\)
\(=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)
Có gì sai sai đấy ạ, cho xin hỏi là có chép sai đề ko ạ?
\(\dfrac{x-2000}{22}\) + \(\dfrac{x-2005}{17}\) + \(\dfrac{x}{674}\) = 5
\(\dfrac{x-2000}{22}\) + \(\dfrac{x-2005}{17}\) + \(\dfrac{x}{674}\) - 5 = 0
(\(\dfrac{x-2000}{22}\) - 1) + (\(\dfrac{x-2005}{17}\) - 1) + (\(\dfrac{x}{674}\) - 3) = 0
\(\dfrac{x-2022}{22}\) + \(\dfrac{x-2022}{17}\) + \(\dfrac{x-2022}{674}\) = 0
(\(x\) - 2022).(\(\dfrac{1}{22}\) + \(\dfrac{1}{17}\) + \(\dfrac{1}{647}\)) = 0
Vì \(\dfrac{1}{22}\) + \(\dfrac{1}{17}\) + \(\dfrac{1}{647}\) > 0
Nên \(x\) - 2022 = 0
\(x\) = 2022
Vậy \(x\) = 2022