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) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)

Giờ mới đúng thật nè

a, (\(\dfrac{1}{3}-\dfrac{1}{2}\))^x-1 = \(\dfrac{1}{36}\)

<=> (\(-\dfrac{1}{6}\))^x-1 = \(\dfrac{1}{36}\)

<=> (\(-\dfrac{1}{6}\))^x-1 = (\(-\dfrac{1}{6}\))²

<=> x - 1 = 2
<=> x = 3
@Nguyễn Thị Phương Thảo

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: \(\Leftrightarrow7^x\cdot49+7^x\cdot\dfrac{2}{7}=345\)

\(\Leftrightarrow7^x=7\)

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) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

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, 1/2xX+3/5xX=-33/25

Xx(1/2+3/5)=-33/25

Xx11/10=-33/25

X=-6/5

b,

\(\dfrac{5^{x+1}}{125}=\dfrac{1}{25^{x-2}}\\ \dfrac{5^{x+1}}{125}=\dfrac{1}{5^{2x-4}}\\ 5^{x+1}\cdot5^{2x-4}=125\\ 5^{x+1+2x-4}=5^3\\ 5^{\left(x+2x\right)+\left(1-4\right)}=5^3\\ 5^{3x-3}=5^3\\ 3x-3=3\\ 3x=6\\ x=2\)

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,73​x−32​x=2110​⇒x(73​−32​)=2110​⇒x.−215​=2110​⇒x=−2b,357​:(x−31​)=−252​⇒51​:(x−31​)=−252​⇒x−31​=−25​⇒x=−613​c,3.(x−21​)−5.(x+53​)=−x+51​⇒3x−23​−5x+5=−x+51​

⇒�(3−5)−32+5=−�+15⇒−2�−132=−�+15⇒−�−135=15⇒�=15−135⇒�=−125.⇒x(3−5)−23​+5=−x+51​⇒−2x−213​=−x+51​⇒−x−513​=51​⇒x=51​−513​⇒x=−512​.

a, \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)

\(\Leftrightarrow\dfrac{2\left(5-x\right)}{2}=\dfrac{5\left(5-x\right)}{5}\)

\(\Leftrightarrow5-x=5-x\)

\(\Leftrightarrow0x=0\)

⇒ Có vô số giá trị của x thỏa mãn.

Vậy...

b, ĐKXĐ: \(x\ne\pm1\)

\(\dfrac{x-3}{x-1}-\dfrac{2x+1}{x+1}=\dfrac{x-x^2}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+1\right)-\left(2x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-x^2}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow x^2-2x-3-2x^2+x+1=x-x^2\)

\(\Leftrightarrow-2x=2\)

\(\Leftrightarrow x=-1\left(ktm\right)\)

Vậy...

a) Ta có: \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)

\(\Leftrightarrow5\left(10-2x\right)=2\left(25-5x\right)\)

\(\Leftrightarrow50-10x=50-10x\)

\(\Leftrightarrow0x=0\)(phương trình có vô số nghiệm)

Vậy: S={x|\(x\in R\)}