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Formula
Solve the equation
Answer
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Graph
$y = \dfrac { 6 } { x }$
$y = \dfrac { \sqrt{ 3 } } { 3 }$
Asymptote
$y = 0$, $x = 0$
Standard form
$y = \dfrac { 6 } { x }$
Domain
$y \neq 0$
Range
$x \neq 0$
$\dfrac{ 6 }{ x } = \dfrac{ \sqrt{ 3 } }{ 3 }$
$x = 6 \sqrt{ 3 }$
Solve the fractional equation
$\color{#FF6800}{ \dfrac { 6 } { x } } = \color{#FF6800}{ \dfrac { \sqrt{ 3 } } { 3 } }$
$ $ If $ \frac{a(x)}{b(x)} = \frac{c(x)}{d(x)} $ is valid, it is $ \begin{cases} a(x) d(x) = b(x) c(x) \\ b(x) \ne 0 \\ d(x) \ne 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ x } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ x } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Simplify the expression $ $
$\begin{cases} \color{#FF6800}{ 18 } = \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ x } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 18 } = \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ x } \\ x \neq 0 \\ 3 \neq 0 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \\ x \neq 0 \\ 3 \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Substitute $ x = 6 \sqrt{ 3 } $ for unresolved equations or inequalities $ $
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} x = 6 \sqrt{ 3 } \\ \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \neq \color{#FF6800}{ 0 } \\ 3 \neq 0 \end{cases}$
$ $ For integers $ a, b, c, n $ , if $ c \ne (-\frac{a}{b})^n $ is valid, it is $ a + b \sqrt[n]{c} \ne 0$
$\begin{cases} x = 6 \sqrt{ 3 } \\ \text{There are countless solutions} \\ 3 \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \\ \text{There are countless solutions} \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Ignore the cases where the system of equations where there are infinitely many solutions. $ $
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} x = 6 \sqrt{ 3 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ There are infinitely many solutions if both sides of $ \ne $ are different. $ $
$\begin{cases} x = 6 \sqrt{ 3 } \\ \text{There are countless solutions} \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } } \\ \text{There are countless solutions} \end{cases}$
$ $ Ignore the cases where the system of equations where there are infinitely many solutions. $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 3 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$13$ Find the roots of the equation $\dfrac {1} {x+1}-\dfrac {1} {x}=3$ 
$A$ $\dfrac {-3-\sqrt{3} } {3},\dfrac {-3+\sqrt{3} } {3}$ $B$ $\dfrac {-2-\sqrt{3} } {3},\dfrac {-2+\sqrt{3} } {3}$ $C$ $\dfrac {-3-\sqrt{3} } {2},\dfrac {-3+\sqrt{3} } {2}$ D. $\dfrac {-1-\sqrt{3} } {3},\dfrac {-1+\sqrt{3} } {3}$
7th-9th grade
Algebra
search-thumbnail-Solve: 

$140$ $\dfrac {\dfrac {1} {3}} {\dfrac {7} {10}}=\dfrac {6} {x}$
10th-13th grade
Other
search-thumbnail-The rationalizing factor of \sqrt{23} is 
$°$ $Options^{°}$ $0$ 
A 24 
23 
C \sqrt{23} 
D None of these
7th-9th grade
Other
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