=>(6/5)^x=(6/5)^3

=>x=3

\(\left(\dfrac{6}{5}\right)^x=\left(\dfrac{6}{5}\right)^3\Leftrightarrow x=3\)

Vậy \(x=3\)

= \(\dfrac{44}{105}\) <  \(\dfrac{x}{210}\) <\(\dfrac{158}{105}\) = \(\dfrac{88}{210}< \dfrac{x}{210}< \dfrac{316}{210}\) 
=> x = 89 -> 315

=> x = 89 -> 315

1.a) Để căn thức có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{2x-1}\ge0\\2x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)

Vậy...

b, \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\dfrac{1}{27}}=\sqrt[3]{\dfrac{625}{5}}-\sqrt[3]{-\dfrac{216}{27}}=\sqrt[3]{125}-\sqrt[3]{-8}=5-\left(-2\right)=7\)

a) Để căn thức có nghĩa thì 2x-1>0

\(\Leftrightarrow2x>1\)

hay \(x>\dfrac{1}{2}\)

b) Ta có: \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}\cdot\sqrt[3]{\dfrac{1}{27}}\)

\(=5-\left(-6\right)\cdot\dfrac{1}{3}\)

\(=5+6\cdot\dfrac{1}{3}=5+2=7\)

Lời giải:

a. $\frac{2-x}{4}=\frac{3x-1}{3}$

$\Rightarrow 3(2-x)=4(3x-1)$

$\Rightarrow 6-3x=12x-4$

$\Rightarrow 6+4=12x+3x$

$\Rightarrow 10=15x$

$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$

b.

$\frac{x}{7}=\frac{x+16}{35}$

$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$

$\Rightarrow 5x=x+16$

$\Rightarrow 4x=16$

$\Rightarrow x=4$

c.

$\sqrt{x^2+1}=3$

$\Rightarrow x^2+1=9$

$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$

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​.

e: =>2/7-x=2/5

=>7-x=5

=>x=2

f: =>2x+3/3=10/3

=>2x+3=10

=>2x=7

=>x=7/2

g: =>(14+x)/7=15/7

=>x+14=15

=>x=1

h: =>(2x+3)/x=13/x

=>2x+3=13

=>2x=10

=>x=5

a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)

    \(\dfrac{3}{7}\)\(x\)         = - \(\dfrac{17}{35}\) + 0,4

     \(\dfrac{3}{7}\)\(x\)       = - \(\dfrac{3}{35}\)

        \(x\)       = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)

        \(x\)       = - \(\dfrac{1}{5}\)

b, 0,2.(\(x\) - 3) +2,4 = 10

    0,2.(\(x\) - 3)          = 10 - 2,4

    0,2.(\(x\) - 3)          = 7,6

           \(x\) - 3            = 7,6:0,2

           \(x\) - 3            = 38

            \(x\)                = 38 + 3

             \(x\)                = 41

\(\dfrac{3}{7}x-0,4=\dfrac{-17}{35}\)

\(\dfrac{3}{7}x=\dfrac{-17}{35}+0,4=\dfrac{-34}{70}+\dfrac{28}{70}\)

\(\dfrac{3}{7}x=\dfrac{-6}{70}=\dfrac{-3}{35}\)

\(x=\dfrac{-3}{35}:\dfrac{3}{7}=\dfrac{-3}{35}\cdot\dfrac{7}{3}\)

\(x=\dfrac{3\cdot\left(-1\right)\cdot7}{5\cdot7\cdot3}=-\dfrac{1}{5}\)

b) \(0,2\left(x-3\right)+2,4=10\)

\(0,2\left(x-3\right)=10-2,4=7,6\)

\(x-3=7,6:0,2=38\)

\(x=38+3=41\)

c) \(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\left(0,5-1\dfrac{3}{5}\right)\)

\(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\)

\(x:2,2=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\cdot\dfrac{6}{1}\)

\(x:2,2=\dfrac{-3\cdot\left(-11\right)\cdot2\cdot3}{2\cdot4\cdot10\cdot1}=\dfrac{99}{40}\)

\(x=\dfrac{99}{40}\cdot2,2=\dfrac{99}{40}\cdot\dfrac{22}{10}\)

\(x=\dfrac{99\cdot2\cdot11}{40\cdot2\cdot5}=\dfrac{1089}{200}\)

\(a,\dfrac{x}{4}=\dfrac{15}{20}\\ \Rightarrow\dfrac{x}{4}=\dfrac{3}{4}\\ \Rightarrow x=3\\ b,\dfrac{15}{x}=\dfrac{25}{35}\\ \Rightarrow\dfrac{15}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{15}{x}=\dfrac{15}{21}\\ \Rightarrow x=21\\ c,\dfrac{x}{5}=\dfrac{26}{65}\\ \Rightarrow\dfrac{x}{5}=\dfrac{2}{5}\\ \Rightarrow x=2\\ d,\dfrac{3}{x}=\dfrac{51}{85}\\ \Rightarrow\dfrac{3}{x}=\dfrac{3}{5}\\ \Rightarrow x=5\)

a,x4=1520⇒x4=34⇒x=3b,15x=2535⇒15x=57⇒15x=1521⇒x=21c,x5=2665⇒x5=25⇒x=2d,3x=5185⇒3x=35⇒x=5