Bài tự luận · Bài 28
Dạng 2. Rút gọn biểu thức chứa luỹ thừa
Bài tự luận Lũy thừa với số mũ thực · Bài 28
(KNTT11) Cho $a$ và $b$ là hai số dương, $a\neq b$. Rút gọn biểu thức sau:
$A=\left[ \frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}} \right]:\left( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right).$
$A=\left[ \frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}} \right]:\left( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right).$
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Lời giải
Vì $\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}=\frac{a-b}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}$ nên
$B=\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}=\frac{a-b}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}=\frac{a-b-a^{\frac{1}{2}}\left( a^{\frac{1}{2}}-b^{\frac{1}{2}} \right)}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}$
$=\frac{a^{\frac{1}{2}}b^{\frac{1}{2}}-b}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}=\frac{b^{\frac{1}{2}}\left( a^{\frac{1}{2}}-b^{\frac{1}{2}} \right)}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}$.
Ta có $a^{\frac{1}{2}}-b^{\frac{1}{2}}=\left( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right)\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)$ 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 $A=\left( \frac{b}{a} \right)^{\frac{1}{2}}\cdot \left( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right)\cdot \frac{1}{a^{\frac{1}{4}}-b^{\frac{1}{4}}}=\left( \frac{b}{a} \right)^{\frac{1}{2}}$.
$B=\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}=\frac{a-b}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}=\frac{a-b-a^{\frac{1}{2}}\left( a^{\frac{1}{2}}-b^{\frac{1}{2}} \right)}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}$
$=\frac{a^{\frac{1}{2}}b^{\frac{1}{2}}-b}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}=\frac{b^{\frac{1}{2}}\left( a^{\frac{1}{2}}-b^{\frac{1}{2}} \right)}{a^{\frac{1}{2}}\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)}$.
Ta có $a^{\frac{1}{2}}-b^{\frac{1}{2}}=\left( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right)\left( a^{\frac{1}{4}}+b^{\frac{1}{4}} \right)$ 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 $A=\left( \frac{b}{a} \right)^{\frac{1}{2}}\cdot \left( a^{\frac{1}{4}}-b^{\frac{1}{4}} \right)\cdot \frac{1}{a^{\frac{1}{4}}-b^{\frac{1}{4}}}=\left( \frac{b}{a} \right)^{\frac{1}{2}}$.