Giải bài tập 6.6 trang 7 SBT toán 11 tập 2 Kết nối

Xét $B=\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}$

$B=\frac{a-b}{a^{\frac{1}{2}}\left ( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right )}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}=\frac{a-b-a^{\frac{1}{2}}\left ( a^{\frac{1}{2}}-b^{\frac{1}{2}} \right )}{a^{\frac{1}{2}}\left ( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right )}$

$B=\frac{a^{\frac{1}{2}}b^{\frac{1}{2}}-b}{a^{\frac{1}{2}}\left ( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right )}$

$B=\frac{b^{\frac{1}{2}}\left ( a^{\frac{1}{2}}-b^{\frac{1}{2}} \right )}{a^{\frac{1}{2}}\left ( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right )}$

Có $a^{\frac{1}{2}}-b^{\frac{1}{2}}=\left ( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right )\cdot \left ( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right )$ nên

$B=\frac{b^{\frac{1}{2}}(a^{\frac{1}{4}}-b^{\frac{1}{4}})(a^{\frac{1}{4}}+b^{\frac{1}{4}})}{a^{\frac{1}{2}}(a^{\frac{1}{4}}+b^{\frac{1}{4}})}=\left ( \frac{b}{a} \right )^{\frac{1}{2}}.(a^{\frac{1}{4}}-b^{\frac{1}{4}})$

=> $A=\left ( \frac{b}{a} \right )^{\frac{1}{2}}.(a^{\frac{1}{4}}-b^{\frac{1}{4}})\cdot \frac{1}{a^{\frac{1}{4}}-b^{\frac{1}{4}}}$

$A=\left ( \frac{b}{a} \right )^{\frac{1}{2}}$