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a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)
\(x=\dfrac{21}{10}\)
\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
\(c,\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)
TH1: \(0,4x-1,3=2,3\)
\(0,4x=3,6\)
\(x=9\)
TH2: \(0,4x-1,3=-2,3\)
\(0,4x=-1\)
\(x=-\dfrac{5}{2}\)
=.= hok tốt!!
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
Ta có: \(\dfrac{\left(x+3\right)^5}{\left(x+3\right)^2}=\dfrac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\dfrac{8}{3}\right)^3\)
\(\Leftrightarrow x+3=\dfrac{8}{3}\)
\(\Leftrightarrow x=\dfrac{8}{3}-3=\dfrac{8}{3}-\dfrac{9}{3}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
\(\dfrac{\left(x+3\right)^5}{\left(x+3\right)^2}=\dfrac{64}{27}\)
\(\Leftrightarrow x+3=\dfrac{4}{3}\)
hay \(x=-\dfrac{5}{3}\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)
\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)
\(\Leftrightarrow376x+752=375x+1125\)
\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)
Ta có: \(3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{15}\)
\(\Leftrightarrow3x-5x+x=\dfrac{1}{15}+\dfrac{3}{2}+3\)
\(\Leftrightarrow x=-\dfrac{137}{30}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
\(\rightarrow\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\rightarrow\dfrac{3}{5}-\dfrac{2}{3}x=\dfrac{-4}{5}\)
\(\rightarrow x=\dfrac{21}{10}\)