12. \( x>0 \) \( \sqrt[3]{16 x^{2}}+\sqrt[2]{25 x^{2}}+\sqrt[2]{36 x^{2}}= \) ?
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Extra Insights
To solve the expression \( \sqrt[3]{16 x^{2}}+\sqrt[2]{25 x^{2}}+\sqrt[2]{36 x^{2}} \), let's evaluate each term separately. 1. For \( \sqrt[3]{16 x^{2}} \): - \( \sqrt[3]{16} = 2\sqrt[3]{2} \) (since \( 16 = 2^4 \)) - Thus, \( \sqrt[3]{16 x^{2}} = \sqrt[3]{16} \cdot \sqrt[3]{x^{2}} = 2\sqrt[3]{2} \cdot x^{\frac{2}{3}} \). 2. For \( \sqrt[2]{25 x^{2}} \): - \( \sqrt{25} = 5 \) and \( \sqrt{x^{2}} = x \) - Thus, \( \sqrt[2]{25 x^{2}} = 5x \). 3. For \( \sqrt[2]{36 x^{2}} \): - \( \sqrt{36} = 6 \) and \( \sqrt{x^{2}} = x \) - Thus, \( \sqrt[2]{36 x^{2}} = 6x \). Now we can combine all three parts: \[ \sqrt[3]{16 x^{2}}+\sqrt[2]{25 x^{2}}+\sqrt[2]{36 x^{2}} = 2\sqrt[3]{2} x^{\frac{2}{3}} + 5x + 6x = 2\sqrt[3]{2} x^{\frac{2}{3}} + 11x. \] So the final expression is: \[ \sqrt[3]{16 x^{2}}+\sqrt[2]{25 x^{2}}+\sqrt[2]{36 x^{2}} = 2\sqrt[3]{2} x^{\frac{2}{3}} + 11x. \]
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