Question

V= ⅓πh (r² +R²+rR) despeje r

necesito despejar r, en esta ecuación ​

Answer 1

PREGUNTA

V= ⅓πh (r² +R²+rR) despeje r

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SOLUCIÓN

۞ HØlα!! ✌

Primero despejaremos el primer factor(este pasará dividiendo)

                           $V = \dfrac{1}{3}\pi h(r^2+R^2 + rR)\\\\\\(r^2+R^2+rR) = \dfrac{3V}{\pi h}\\\\\\r^2+rR =\dfrac{3V}{\pi h}-R^2\\\\\\\mathrm{Sumamos \: y \: restamos (R^2/4)}\\\\\\\underbrace{r^2+rR + \boldsymbol{\dfrac{R^2}{4}}}_{TCP}\boldsymbol{-\dfrac{R^2}{4}}=\dfrac{3V}{\pi h}-R^2\\\\\\TCP: Trinomio \: cuadrado \: perfecto\\\\\\\left(r+\dfrac{R}{2}\right)^2 -\dfrac{R^2}{4} = \dfrac{3V}{\pi h}-R^2\\\\\\\left(r+\dfrac{R}{2}\right)^2 = \left(\dfrac{3V}{\pi h}-R^2\right) +\dfrac{R^2}{4}$

                           $\left(r+\dfrac{R}{2}\right)^2 = \left(\dfrac{3V}{\pi h}-R^2\right) +\dfrac{R^2}{4}\\\\\\r+\dfrac{R}{2} = \pm\sqrt{\left(\dfrac{3V}{\pi h}-R^2\right) +\dfrac{R^2}{4}}\\\\\\\boxed{\boxed{\boldsymbol{r = -\dfrac{R}{2} \pm\sqrt{\left(\dfrac{3V}{\pi h}-R^2\right) +\dfrac{R^2}{4}}}}}$