Giải bài 8 trang 78 SGK giải tích nâng cao 12
Đơn giản biểu thức
a) \(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{a}-\sqrt[4]{b}}-\dfrac{\sqrt{a}+\sqrt[4]{ab}}{\sqrt[4]{a}+\sqrt[4]{b}}\); b) \(\dfrac{a-b}{\sqrt[3]{a}-\sqrt[3]{b}}-\dfrac{a+b}{\sqrt[3]{a}+\sqrt[3]{b}}\);
c) \( \left( \dfrac{a+b}{\sqrt[3]{a}+\sqrt[3]{b}}-\sqrt[3]{ab} \right):{{\left( \sqrt[3]{a}-\sqrt[3]{b} \right)}^{2}} \); d) \(\dfrac{a-1}{{{a}^{\frac{3}{4}}}+{{a}^{\frac{1}{2}}}}.\dfrac{\sqrt{a}+\sqrt[4]{a}}{\sqrt{a}+1}.{{a}^{\frac{1}{4}}}+1 \).
\(\begin{align} &a)\, \dfrac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{a}-\sqrt[4]{b}}-\dfrac{\sqrt{a}+\sqrt[4]{ab}}{\sqrt[4]{a}+\sqrt[4]{b}} \\ & =\sqrt[4]{a}+\sqrt[4]{b}-\dfrac{\sqrt[4]{a}\left( \sqrt[4]{a}+\sqrt[4]{b} \right)}{\sqrt[4]{a}+\sqrt[4]{b}} \\ & =\sqrt[4]{a}+\sqrt[4]{b}-\sqrt[4]{a} \\ & =\sqrt[4]{b} \\ \end{align} \)
\(\begin{align} &b)\, \dfrac{a-b}{\sqrt[3]{a}-\sqrt[3]{b}}-\dfrac{a+b}{\sqrt[3]{a}+\sqrt[3]{b}} \\ & =\sqrt[3]{{{a}^{2}}}+\sqrt[3]{ab}+\sqrt[3]{{{b}^{2}}}-\left( \sqrt[3]{{{a}^{2}}}-\sqrt[3]{ab}+\sqrt[3]{{{b}^{2}}} \right) \\ & =2\sqrt[3]{ab} \\ \end{align} \)
\(\begin{align} &c)\, \left( \dfrac{a+b}{\sqrt[3]{a}+\sqrt[3]{b}}-\sqrt[3]{ab} \right):{{\left( \sqrt[3]{a}-\sqrt[3]{b} \right)}^{2}} \\ & =\left( \sqrt[3]{{{a}^{2}}}-\sqrt[3]{ab}+\sqrt[3]{{{b}^{2}}}-\sqrt[3]{ab} \right):{{\left( \sqrt[3]{a}-\sqrt[3]{b} \right)}^{2}} \\ & ={{\left( \sqrt[3]{a}-\sqrt[3]{b} \right)}^{2}}:{{\left( \sqrt[3]{a}-\sqrt[3]{b} \right)}^{2}} \\ & =1 \\ \end{align} \)
\(\begin{align} &d)\, \dfrac{a-1}{{{a}^{\frac{3}{4}}}+{{a}^{\frac{1}{2}}}}.\dfrac{\sqrt{a}+\sqrt[4]{a}}{\sqrt{a}+1}.{{a}^{\frac{1}{4}}}+1 \\ & =\dfrac{\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}{\sqrt{a}\left( \sqrt[4]{a}+1 \right)}.\dfrac{\sqrt[4]{a}\left( \sqrt[4]{a}+1 \right)}{\sqrt{a}+1}.{{a}^{\frac{1}{4}}}+1 \\ & =\sqrt{a}-1+1 \\ & =\sqrt{a} \\ \end{align} \)