Đáp án: $14:B;15:C$
Giải thích các bước giải:
Câu 14:
$\large(\frac{a+b}{\sqrt[3]{a}+\sqrt[3]{b}}-\sqrt[3]{ab})÷(\sqrt[3]{a}-\sqrt[3]{b})^2$
$\large=(\frac{(\sqrt[3]{a}+\sqrt[3]{b})[(\sqrt[3]{a})^2-\sqrt[3]{a}.\sqrt[3]{b}+(\sqrt[3]{b})^2]}{\sqrt[3]{a}+\sqrt[3]{b}}-\sqrt[3]{a}.\sqrt[3]{b})÷(\sqrt[3]{a}-\sqrt[3]{b})^2$
$\large=[(\sqrt[3]{a})^2-\sqrt[3]{a}.\sqrt[3]{b}+(\sqrt[3]{b})^2-\sqrt[3]{a}.\sqrt[3]{b}]÷(\sqrt[3]{a}-\sqrt[3]{b})^2$
$\large=[(\sqrt[3]{a})^2-2\sqrt[3]{a}.\sqrt[3]{b}+(\sqrt[3]{b})^2]÷(\sqrt[3]{a}-\sqrt[3]{b})^2$
$\large=(\sqrt[3]{a}-\sqrt[3]{b})^2÷(\sqrt[3]{a}-\sqrt[3]{b})^2=1$
Câu 15:
`a^\frac{5}{2}(a>0)`
`=a^{5.\frac{1}{2}}=(a^5)^\frac{1}{2}=\sqrt{a^5}`
$=a^2\sqrt{a}=\sqrt[3]{a^6}\sqrt{a}$
$=\large\frac{\sqrt[3]{a^7}}{\sqrt[3]{a}}.\sqrt{a}=\large\frac{\sqrt[3]{a^7}\sqrt{a}}{\sqrt[3]{a}}$
