Cho a và b là số dương, a khác b. Rút gọn biểu thức sau, bài 6.6 trang 7 SBT

Giải SBT Toán 11 Bài 18: Lũy thừa với số mũ thực

Bài 6.6 trang 7 SBT Toán 11 Tập 2: Cho a và b là số dương, a ≠ b. Rút gọn biểu thức sau:

$A=\left[\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}{b}^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)$ .

Lời giải:

Đặt $B=\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}{b}^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}=\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}$

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

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

Do đó

$\mathrm{A}=\left[\frac{\mathrm{a}-\mathrm{b}}{{\mathrm{a}}^{\frac{3}{4}}+{\mathrm{a}}^{\frac{1}{2}}{\mathrm{b}}^{\frac{1}{4}}}-\frac{{\mathrm{a}}^{\frac{1}{2}}-{\mathrm{b}}^{\frac{1}{2}}}{{\mathrm{a}}^{\frac{1}{4}}+{\mathrm{b}}^{\frac{1}{4}}}\right]:\left({\mathrm{a}}^{\frac{1}{4}}-{\mathrm{b}}^{\frac{1}{4}}\right)$$=\left[{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\frac{1}{2}}\left({\mathrm{a}}^{\frac{1}{4}}-{\mathrm{b}}^{\frac{1}{4}}\right)\right]:\left({\mathrm{a}}^{\frac{1}{4}}-{\mathrm{b}}^{\frac{1}{4}}\right)$

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