Bài 1: Lũy thừa

Lý thuyết

Mục lục

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Bài 3 (SGK trang 56)

Viết các số sau theo thứ tự tăng dần:

$a)\ 1^{3,75};\ 2^{-1};\ (\dfrac{1}{2})^{-3}$

$b)\ 98^0;\ (\dfrac{3}{7})^{-1};\ 32^{\dfrac{1}{5}}$

Hướng dẫn giải

a) $1^{3,75}$ = 1 = $2^{0}$ ; $\left ( \frac{1}{2} \right )^{-3}$ = $2^{3}$ .

Mặt khác trong hai lũy thừa cungc cơ số lớn hơn 1, lũy thừa nào có số mũ lớn hơn là lũy thừa lớn hơn. Do đó theo thứ tự tăng dần ta được:

$2^{-1}$ < $1^{3,75}$ < $\left ( \frac{1}{2} \right )^{-3}$

b) $98^{0}$ = 1 = $\frac{3}{3}$ ; $\left ( \frac{3}{7} \right )^{-1}$ = $\frac{7}{3}$ ; $32^{\frac{1}{5}}$ = $\left ( 2^{5} \right )^{\frac{1}{5}}$ = 2 = $\frac{6}{3}$ .

Do đó $98^{0}$ < $32^{\frac{1}{5}}$ < $\left ( \frac{3}{7} \right )^{-1}$ .

Bài 1 (SGK trang 55)

Tính:

$a)\ 9^{\dfrac{2}{5}}.27^{\dfrac{2}{5}}$

$b)\ 144^{\dfrac{3}{4}}:9^{\dfrac{3}{4}}$

$c)\ (\dfrac{1}{16})^{-0,75}+(0,25)^{\dfrac{-5}{2}}$

$d)\ (0,04)^{-1,5}-(0,125)^{\dfrac{-2}{3}} $

Hướng dẫn giải

a) $9^{\frac{2}{5}}$ . $27^{\frac{2}{5}}$ = $\left ( 9.27 \right )^{\frac{2}{5}}$ = $\left ( 3^{2}.3^{3} \right )^{\frac{2}{5}}$ = $\left ( 3^{5\frac{2}{5}} \right )$ = $3^{2}$ = 9.

b) $144^{\frac{3}{4}}$ : $9^{\frac{3}{4}}$ = $\left ( 144:9^{\frac{3}{4}} \right )$ = $\left ( \left ( \frac{12}{3} \right )^{2} \right )^{\frac{3}{4}}$ = $\left ( 4^{2\frac{3}{4}} \right )$ = $4^{\frac{3}{2}}$ = $2^{3}$ = 8.

c) $\left ( \frac{1}{16} \right )^{-0,75}$ + $\left ( 0,25 \right )^{\frac{-5}{2}}$ = $16^{0,75}$ + $\left ( \frac{1}{4} \right )^{\frac{-5}{2}}$ = $\left ( 2^{4} \right )^{0,75}$ + $4^{2,5}$ = $2^{4.0,75}$ + $2^{2.2,5}$ = $2^{3}$ + $2^{5}$ = 40.

d) $\left ( 0,04 \right )^{-1,5}$ - $\left ( 0,125 \right )^{\frac{-2}{3}}$ = $\left ( \frac{4}{100} \right )^{-1,5}$ - $\left ( \frac{125}{1000} \right )^{\frac{-2}{3}}$ = $\left ( \frac{100}{4} \right )^{1,5}$ - $8^{\frac{2}{3}}$ = $5^{2. 1,5}$ - $\left 2^{3\frac{2}{3}} \right$ = 121.

Bài 4 (SGK trang 56)

Cho a, b là những số thực dương. Rút gọn các biểu thức sau:

$a)\ \dfrac{a^{\dfrac{4}{3}}(a^{\dfrac{-1}{3}}+a^{\dfrac{2}{3}})}{a^{\dfrac{1}{4}}(a^{\dfrac{3}{4}}+a^{\dfrac{-1}{4}})}$

$b)\ \dfrac{b^{\dfrac{1}{5}} (\sqrt[5]{b^4}-\sqrt[5]{b^{-1}})}{b^{\dfrac{2}{3}}(\sqrt[3]{b}-\sqrt[3]{b^{-2}})}$

$c)\ \dfrac{a^{\dfrac{1}{3}}b^{\dfrac{-1}{3}}-a^{\dfrac{-1}{3}}b^{\dfrac{1}{3}}} {\sqrt[3]{a^2}-\sqrt[3]{b^2}}$
$d)\ \dfrac{a^{\dfrac{1}{3}} \sqrt{b}+b^{\dfrac{1}{3}} \sqrt{a}} {\sqrt[6]{a}+\sqrt[6]{b}}$

Hướng dẫn giải

a) $\frac{a^{\frac{4}{3}}\left ( a^{\frac{-1}{3}}+ a^{\frac{2}{3}} \right )}{a^{\frac{1}{4}}\left ( a^{\frac{3}{4}}+ a^{\frac{-1}{4}} \right )}$ = $\frac{a^{\frac{4}{3}-\frac{1}{3}} + a^{\frac{4}{3}+\frac{2}{3}} }{a^{\frac{1}{4}+\frac{3}{4}}+a^{\frac{1}{4}+\frac{-1}{4}}}$ = $\frac{a^{1}+a^{2}}{a^{1}+a^{0}}= \frac{a\left ( 1+a \right )}{a+1}= a$

b) $\frac{b^{\frac{1}{5}}\left ( \sqrt[5]{b^{4}}- \sqrt[5]{b^{-1}} \right )}{b^{\frac{2}{3}}\left (\sqrt[3]{b}- \sqrt[3]{b^{-2}} \right )} ;$ = $\frac{b^{\frac{1}{5}}\left (b ^{\frac{4}{5}}-b^{\frac{-1}{5}} \right )}{b^{\frac{2}{3}}\left ( b^{\frac{1}{3}}-b^{\frac{-2}{3}} \right )}$ = $\frac{b^{\frac{1}{5}-\frac{4}{5}} - b^{\frac{1}{5}-\frac{1}{5}} }{b^{\frac{2}{3}+\frac{1}{3}}-b^{\frac{2}{3}+\frac{2}{3}}}$ = $\frac{b-1}{b-1}= 1$ . ( Với điều kiện b # 1)

c) $\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}$= $\frac{a^{\frac{-1}{3}}b^{\frac{-1}{3}}\left ( a^{\frac{2}{3}}-b^{\frac{2}{3}}\right )}{a^{\frac{2}{3}}-b^{\frac{2}{3}}}$ = $a^{\frac{-1}{3}}b^{\frac{-1}{3}}$ = $\frac{1}{a^{\frac{1}3{}}b^{\frac{1}{3}}}= \frac{1}{\sqrt[3]{ab}}$ ( với điều kiện a#b).

d) $\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}$ = $\frac{a^{\frac{1}{3}}b^{\frac{1}{2}}+b^{\frac{1}{3}}a^{\frac{1}{2}}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}$ = $\frac{a^{\frac{2}{6}}b^{\frac{3}{6}}+b^{\frac{2}{6}}a^{\frac{3}{6}}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}$ = $\frac{a^{\frac{2}{6}}b^{\frac{2}{6}}\left ( a^{\frac{1}{6}}+b^{\frac{1}{6}}\right )}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}$ = $a^{\frac{2}{6}}b^{\frac{2}{6}} = a^{\frac{1}{3}}b^{\frac{1}{3}} = \sqrt[3]{ab}.$