Giải bài 6.6 trang 7

Đề bài

Cho $a$ và $b$ là hai số dương, $a\ne b$. 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 chi tiết

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

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

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

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

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

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