tim x : a, (\(\dfrac{1}{3}\)-\(\dfrac{1}{2}\))\(^{x-1}\)=\(\dfrac{1}{36}\)
b,\(\dfrac{25}{5^x}\)=\(\dfrac{1}{125}\)
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a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^x-1=\dfrac{1}{36}\)
\(\Rightarrow\left(\dfrac{1}{6}\right)^x=1\dfrac{1}{36}\)
\(\Rightarrow x=2\)
b) \(\dfrac{25}{5x}=\dfrac{1}{125}\)
\(\Rightarrow25\cdot125=5x\)
\(\Rightarrow5^2\cdot5^3=5^x\)
\(\Rightarrow x=5\)
Bn Đời về cơ bản là buồn... cười!!! nhìn sai đề ở ý a .
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\dfrac{1}{6}^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
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. \(\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{-33}{25}\)
\(\Rightarrow\dfrac{11}{10}x=\dfrac{-33}{25}\)
\(\Rightarrow x=\dfrac{-33}{25}:\dfrac{11}{10}=\dfrac{-6}{5}\)
Vậy.........
b. \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}+\dfrac{-3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\\dfrac{1}{2}+\dfrac{-3}{7}:x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{-3}{7}:x=\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{6}{7}\end{matrix}\right.\)
Vậy................
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
a: \(\Leftrightarrow7^x\cdot49+7^x\cdot\dfrac{2}{7}=345\)
\(\Leftrightarrow7^x=7\)
hay x=1
c: \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
=>x-1=2
hay x=3
d: \(\dfrac{25}{5^x}=\dfrac{1}{125}\)
\(\Leftrightarrow5^x=5^2\cdot5^3=5^5\)
hay x=5
a: \(\Leftrightarrow x^2=900\)
=>x=30 hoặc x=-30
b: \(\Leftrightarrow\dfrac{2}{3}:\left(-0.1x\right)=\dfrac{4}{3}:\dfrac{-2}{25}=-\dfrac{4}{3}\cdot\dfrac{25}{2}=-\dfrac{100}{6}=\dfrac{-50}{3}\)
=>0,1x=2/3:50/3=2/3x3/50=1/25
=>1/10x=1/25
hay x=1/25:1/10=10/25=2/5
d: \(\Leftrightarrow x^2=\dfrac{144}{25}\)
=>x=12/5 hoặc x=-12/5
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\) \(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\left(\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=2+1=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
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