Question

Resolva a equação:

$\sqrt{ \frac{2x}{1-3x}} + \sqrt{ \frac{1-3x}{2x}} = \frac{13}{6}$

Answer 1

Para simplificar, faça a substituição $y=\dfrac{2x}{1-3x}$

Com isso, $\dfrac{1-3x}{2x}=\dfrac{1}{y}$

Além disso, podemos achar x em função de y:

$\dfrac{2x}{1-3x}=y\,\,\Leftrightarrow\,\,2x=y-3xy\,\,\Leftrightarrow\,\,x(2+3y)=y\,\,\Leftrightarrow\,\,x=\dfrac{y}{2+3y}$

Note que, para as raízes estarem bem definidas, devemos ter

$\mathsf{\bullet\,\,2x\ge0\,\,\,\,e\,\,\,\,1-3x\neq0\,\,\,\Longleftrightarrow\,\,\,x\ge0\,\,\,\,e\,\,\,\,x\neq\frac{1}{3}}\\\\\mathsf{\bullet\,\,1-3x\ge0\,\,\,\,e\,\,\,\,2x\neq0\,\,\,\Longleftrightarrow\,\,\,x\le\frac{1}{3}\,\,\,\,e\,\,\,\,x\neq0}$

Como todas essas condições devem ser satisfeitas, $x\in(0,\frac{1}{3})$.
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Resolvendo a equação:

$\sqrt{\dfrac{2x}{1-3x}}+\sqrt{\dfrac{1-3x}{2x}}=\dfrac{13}{6}\\\\\\\sqrt{y}+\sqrt{\dfrac{1}{y}}=\dfrac{13}{6}$

Elevando os dois lados da igualdade ao quadrado:

$\bigg(\sqrt{y}+\sqrt{\dfrac{1}{y}}\bigg)^{2}=\bigg(\dfrac{13}{6}\bigg)^{2}\\\\\\\big(\sqrt{y}\big)^{2}+2\sqrt{y}\cdot\sqrt{\dfrac{1}{y}}+\bigg(\sqrt{\dfrac{1}{y}}\bigg)^{2}=\dfrac{169}{36}\\\\\\y+2\sqrt{y\cdot\dfrac{1}{y}}+\dfrac{1}{y}=\dfrac{169}{36}\\\\\\y+2\sqrt{1}+\dfrac{1}{y}=\dfrac{169}{36}\\\\\\y+2+\dfrac{1}{y}=\dfrac{169}{36}\\\\\\y+\dfrac{1}{y}=\dfrac{169}{36}-2=\dfrac{169}{36}-\dfrac{72}{36}\\\\\\y+\dfrac{1}{y}=\dfrac{97}{36}$

Multiplicando os dois lados da equação por y:

$y\bigg(y+\dfrac{1}{y}\bigg)=\dfrac{97}{36}y\\\\\\y^{2}+1=\dfrac{97}{36}y\\\\\\y^{2}-\dfrac{97}{36}y+1=0$

Resolvendo por Bhaskara:

$\Delta=b^{2}-4ac\\\\\Delta=\big(-\frac{97}{36}\big)^{2}-4\cdot1\cdot1\\\\\Delta=\frac{9409}{1296}-4=\frac{9409}{1296}-\frac{5184}{1296}\\\\\Delta=\frac{4225}{1296}\\\\\Delta=\frac{65^{2}}{36^{2}}=\big(\frac{65}{36}\big)^{2}$

Com isso, $\sqrt{\Delta}=\sqrt{\big(\frac{65}{36}\big)^{2}}=\frac{65}{36}$

$y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-\big(-\frac{97}{36}\big)\pm\frac{65}{36}}{2\cdot1}=\dfrac{97\pm65}{2\cdot36}\begin{cases}y=\dfrac{97+65}{72}=\dfrac{162}{72}=\dfrac{81}{36}\\\\\\y=\dfrac{97-65}{72}=\dfrac{32}{72}=\dfrac{16}{36}\end{cases}$
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Caso $y=\frac{81}{36}$:

$x=\dfrac{y}{2+3y}\\\\\\x=\dfrac{\big(\frac{81}{36}\big)}{2+3\big(\frac{81}{36}\big)}\\\\\\x=\dfrac{\big(\frac{81}{36}\big)}{\big(\frac{72}{36}\big)+\big(\frac{243}{36}\big)}\\\\\\x=\dfrac{\big(\frac{81}{36}\big)}{\big(\frac{315}{36}\big)}\\\\\\x=\dfrac{81}{36}\cdot\dfrac{36}{315}\\\\\\x=\dfrac{81}{315}\\\\\\\boxed{x=\dfrac{9}{35}}$

Caso $y=\frac{16}{36}$:

$x=\dfrac{y}{2+3y}\\\\\\x=\dfrac{\big(\frac{16}{36}\big)}{2+3\big(\frac{16}{36}\big)}\\\\\\x=\dfrac{\big(\frac{16}{36}\big)}{\big(\frac{72}{36}\big)+\big(\frac{48}{36}\big)}\\\\\\x=\dfrac{\big(\frac{16}{36}\big)}{\big(\frac{120}{36}\big)}\\\\\\x=\dfrac{16}{120}\\\\\\\boxed{x=\dfrac{2}{15}}$
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Note que ambas são soluções válidas, pois

$\bullet\,\,\dfrac{9}{35}\,\,\textless\,\,\dfrac{1}{3}\,\,\,\Longleftrightarrow\,\,\,9\cdot3\,\,\textless\,\,1\cdot35\,\,\,\Longleftrightarrow\,\,\,27\,\,\textless\,\,35\,\,\,\,(\checkmark)\\\\\\\bullet\,\,\dfrac{2}{15}\,\,\textless\,\,\dfrac{1}{3}\,\,\,\Longleftrightarrow\,\,\,2\cdot3\,\,\textless\,\,1\cdot15\,\,\,\Longleftrightarrow\,\,\,6\,\,\textless\,\,15\,\,\,\,(\checkmark)$

Com isso, o conjunto solução dessa equação é

$\boxed{\boxed{S=\bigg\{\dfrac{2}{15},\,\dfrac{9}{35}\bigg\}}}$

Answer 2

Temos: 

$\sqrt{\frac{2x}{1-3x}}+\sqrt{\frac{1-3x}{2x}}=\frac{13}{6}$

Elevando ambos os lados ao quadrado ficamos com:

$\frac{2x}{1-3x}+2.\sqrt{\frac{2x}{1-3x}.\frac{1-3x}{2x}}+\frac{1-3x}{2x}=\frac{169}{36}\\ \\\frac{2x}{1-3x}+2+\frac{1-3x}{2x}=\frac{169}{36}\\ \\\frac{2x}{1-3x}+\frac{1-3x}{2x}=\frac{169}{36}-2$

Para simplificar a análise considera-se o seguinte:

$\frac{2x}{1-3x}=T$

Substitui-se:

$T+\frac{1}{T}=\frac{97}{36}\\ \\T^{2}-T\frac{97}{36}+1=0$

O que, pela fórmula de Bhaskara, implica em:

$\sqrt{\Delta}=\frac{65}{36}$

Então:

$T_1=(\frac{97}{36}+\frac{65}{36}).\frac{1}{2}\to T_1=\frac{81}{36}=\boxed{\frac{9}{4}}\\ \\T_2=(\frac{97}{36}-\frac{65}{36}).\frac{1}{2} \to T_2=\frac{16}{36}=\boxed{\frac{4}{9}}$

Como $T=\frac{2x}{1-3x}$, faremos:

 Para $\bold{T_1}$

$\frac{2x}{1-3x}=\frac{9}{4}\\ \\8x=9-27x\\ \\35x=9\\ \\\boxed{x=\frac{9}{35}}$

Para $\bold{T_2}$

$\frac{2x}{1-3x}=\frac{4}{9}\\ \\18x=4-12x\\ \\30x=4 \\ \\ x=\frac{4}{30}=\boxed{\frac{2}{15}}$

Temos como solução:

$\boxed{\boxed{x=\frac{9}{35}, x=\frac{2}{15}}}$