Se da la desigualdad:
x 3 x − 5 ≤ 5 x \sqrt{3 x - 5} \leq 5 x 3 x − 5 ≤ 5 Para resolver esta desigualdad primero hay que resolver la ecuación correspondiente:
x 3 x − 5 = 5 x \sqrt{3 x - 5} = 5 x 3 x − 5 = 5 Resolvemos:
x 1 = 25 81 25 789 162 + 6325 1458 3 + 5 9 + 25 789 162 + 6325 1458 3 x_{1} = \frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{5}{9} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}} x 1 = 81 3 162 25 789 + 1458 6325 25 + 9 5 + 3 162 25 789 + 1458 6325 x 1 = 25 81 25 789 162 + 6325 1458 3 + 5 9 + 25 789 162 + 6325 1458 3 x_{1} = \frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{5}{9} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}} x 1 = 81 3 162 25 789 + 1458 6325 25 + 9 5 + 3 162 25 789 + 1458 6325 Las raíces dadas
x 1 = 25 81 25 789 162 + 6325 1458 3 + 5 9 + 25 789 162 + 6325 1458 3 x_{1} = \frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{5}{9} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}} x 1 = 81 3 162 25 789 + 1458 6325 25 + 9 5 + 3 162 25 789 + 1458 6325 son puntos de cambio del signo de desigualdad en las soluciones.
Primero definámonos con el signo hasta el punto extremo izquierdo:
x 0 ≤ x 1 x_{0} \leq x_{1} x 0 ≤ x 1 Consideremos, por ejemplo, el punto
x 0 = x 1 − 1 10 x_{0} = x_{1} - \frac{1}{10} x 0 = x 1 − 10 1 =
− 1 10 + ( 25 81 25 789 162 + 6325 1458 3 + 5 9 + 25 789 162 + 6325 1458 3 ) - \frac{1}{10} + \left(\frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{5}{9} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}\right) − 10 1 + 81 3 162 25 789 + 1458 6325 25 + 9 5 + 3 162 25 789 + 1458 6325 =
25 81 25 789 162 + 6325 1458 3 + 41 90 + 25 789 162 + 6325 1458 3 \frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{41}{90} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}} 81 3 162 25 789 + 1458 6325 25 + 90 41 + 3 162 25 789 + 1458 6325 lo sustituimos en la expresión
x 3 x − 5 ≤ 5 x \sqrt{3 x - 5} \leq 5 x 3 x − 5 ≤ 5 − 5 + 3 ( 25 81 25 789 162 + 6325 1458 3 + 41 90 + 25 789 162 + 6325 1458 3 ) ( 25 81 25 789 162 + 6325 1458 3 + 41 90 + 25 789 162 + 6325 1458 3 ) ≤ 5 \sqrt{-5 + 3 \left(\frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{41}{90} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}\right)} \left(\frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{41}{90} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}\right) \leq 5 − 5 + 3 81 3 162 25 789 + 1458 6325 25 + 90 41 + 3 162 25 789 + 1458 6325 81 3 162 25 789 + 1458 6325 25 + 90 41 + 3 162 25 789 + 1458 6325 ≤ 5 __________________________________________________________________
/ ___________________ / ___________________ \
/ / _____ | / _____ |
/ 109 / 6325 25*\/ 789 25 |41 / 6325 25*\/ 789 25 |
/ - --- + 3*3 / ---- + ---------- + --------------------------- *|-- + 3 / ---- + ---------- + ---------------------------|
/ 30 \/ 1458 162 ___________________ |90 \/ 1458 162 ___________________| <= 5
/ / _____ | / _____ |
/ / 6325 25*\/ 789 | / 6325 25*\/ 789 |
/ 27*3 / ---- + ---------- | 81*3 / ---- + ---------- |
\/ \/ 1458 162 \ \/ 1458 162 /
significa que la solución de la desigualdad será con:
x ≤ 25 81 25 789 162 + 6325 1458 3 + 5 9 + 25 789 162 + 6325 1458 3 x \leq \frac{25}{81 \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}}} + \frac{5}{9} + \sqrt[3]{\frac{25 \sqrt{789}}{162} + \frac{6325}{1458}} x ≤ 81 3 162 25 789 + 1458 6325 25 + 9 5 + 3 162 25 789 + 1458 6325 _____
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