Câu 1: Trang 55- sgk giải tích 12
Tính:
a) $9^{\frac{2}{5}}.27^{\frac{2}{5}}$
b) $144^{\frac{3}{4}}.9^{\frac{3}{4}}$
c) $(\frac{1}{16})^{-0,75}+0,25^{\frac{-5}{2}}$
d) $(0,04)^{-1,5}-(0,125)^{-\frac{2}{3}}$
Hướng dẫn giải:
Áp dụng các tính chất của hàm lũy thừa, ta có:
a) $9^{\frac{2}{5}}.27^{\frac{2}{5}}$
= $(9.27)^{\frac{2}{5}}$
= $(3^{2}.3^{3})^{\frac{2}{5}}$
= $(3^{2+3})^{\frac{2}{5}}$
= $(3^5)^{\frac{2}{5}}$
= $3^{2}=9$
Vậy $9^{\frac{2}{5}}.27^{\frac{2}{5}}=9$
b) $144^{\frac{3}{4}}.9^{\frac{3}{4}}$
= $(\frac{144}{9})^{\frac{3}{4}}$
= $16^{\frac{3}{4}}$
= $(2^{4})^{\frac{3}{4}}$
= $2^{3}=8$
Vậy $144^{\frac{3}{4}}.9^{\frac{3}{4}}=8$
c) $(\frac{1}{16})^{-0,75}+0,25^{\frac{-5}{2}}$
= $16^{0,75}+(\frac{1}{4})^{-\frac{5}{2}}$
= $16^{\frac{3}{4}}+4^{\frac{5}{2}}$
= $(2^{4})^{\frac{3}{4}}+(2^{2})^{\frac{5}{2}}$
= $2^{3}+2^{5}=8+32=40$
Vậy $(\frac{1}{16})^{-0,75}+0,25^{\frac{-5}{2}}=40$
d) $(0,04)^{-1,5}-(0,125)^{-\frac{2}{3}}$
= $(\frac{4}{100})^{-\frac{3}{2}}-(\frac{1}{8})^{-\frac{2}{3}}$
= $(\frac{100}{4})^{\frac{2}{3}}-8^{\frac{2}{3}}$
= $(5^{2})^{\frac{2}{3}}-(2^{3})^{\frac{2}{3}}$
= $5^{3}-2^{2}=125-4=121$
Vậy $(0,04)^{-1,5}-(0,125)^{-\frac{2}{3}}=121$
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Câu 2: Trang 55- sgk giải tích 12
Viết các biểu thức sau dưới dạng lũy thừa với số mũ hữu tỉ:
a) $a^{\frac{1}{3}}.\sqrt{a}$
b) $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}$
c) $a^{\frac{4}{3}}:\sqrt[3]{a}$
d) $\sqrt[3]{b}:b^{\frac{1}{6}}$
Hướng dẫn giải:
Áp dụng công thức lũy thừa, ta có:
a) $a^{\frac{1}{3}}.\sqrt{a}$
= $a^{\frac{1}{3}}.a^{\frac{1}{2}}$
= $a^{\frac{1}{3}+\frac{1}{2}}=a^{\frac{5}{6}}$
Vậy $a^{\frac{1}{3}}.\sqrt{a}=a^{\frac{5}{6}}$
b) $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}$
= $b^{\frac{1}{2}}.b^{\frac{1}{3}}.b^{\frac{1}{6}}$
= $b^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=b$
Vậy $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}=b$
c) $a^{\frac{4}{3}}:\sqrt[3]{a}$
= $a^{\frac{4}{3}}:a^{\frac{1}{3}}$
= $a^{\frac{4}{3}-\frac{1}{3}}=a$
Vậy $a^{\frac{4}{3}}:\sqrt[3]{a}=a$
d) $\sqrt[3]{b}:b^{\frac{1}{6}}$
= $b^{\frac{1}{3}}:b^{\frac{1}{6}}$
= $b^{\frac{1}{3}-\frac{1}{6}}=b^{\frac{1}{6}}$
Vậy $\sqrt[3]{b}:b^{\frac{1}{6}}=b^{\frac{1}{6}}$
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Câu 3: Trang 56- sgk giải tích 12
Viết các số sau theo thứ tự tăng dần:
a) $1^{3,75};2^{-1};(\frac{1}{2})^{-3}$
b) $98^{0};(\frac{3}{7})^{-1};32^{\frac{1}{5}}$
Hướng dẫn giải:
Ta có:
a) $1^{3,75}=1$
$2^{-1}=\frac{1}{2}$
$(\frac{1}{2})^{-3}=2^{3}$
=> Thứ tự sắp xếp tăng dần: $2^{-1}<1^{3,75}<(\frac{1}{2})^{-3}$
b) $98^{0}=1$
$(\frac{3}{7})^{-1}=\frac{7}{7}$
$32^{\frac{1}{5}}=2$
=> Thứ tự sắp xếp tăng dần: $98^{0}<32^{\frac{1}{5}}<(\frac{3}{7})^{-1}$
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Câu 4: Trang 56- sgk giải tích 12
Rút gọn các biểu thức sau:
a) $\frac{a^{\frac{4}{3}}(a^{-\frac{1}{3}}+a^{\frac{2}{3}})}{a^{\frac{1}{4}}(a^{\frac{3}{4}}+a^{-\frac{1}{4}})}$
b) $\frac{b^{\frac{1}{5}}(\sqrt[5]{b^{4}}-\sqrt[5]{b^{-1}})}{b^{\frac{2}{3}}(\sqrt[3]{b}-\sqrt[3]{b^{-2}})}$
c) $\frac{a^{\frac{1}{3}}b^{-\frac{1}{3}}-a^{-\frac{1}{3}}b^{\frac{1}{3}}}{\sqrt[3]{a^{2}}-\sqrt[3]{b^{2}}}$
d) $\frac{a^{\frac{1}{3}}\sqrt{b}+b^{\frac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}$
Hướng dẫn giải:
a) $\frac{a^{\frac{4}{3}}(a^{-\frac{1}{3}}+a^{\frac{2}{3}})}{a^{\frac{1}{4}}(a^{\frac{3}{4}}+a^{-\frac{1}{4}})}$
= $\frac{a^{\frac{4}{3}}.a^{-\frac{1}{3}}+a^{\frac{4}{3}}.a^{\frac{2}{3}}}{a^{\frac{1}{4}}.a^{\frac{3}{4}}+a^{\frac{1}{4}}.a^{-\frac{1}{4}}}$
= $\frac{a^{\frac{4}{3}-\frac{1}{3}}+a^{\frac{4}{3}+\frac{2}{3}}}{a^{\frac{1}{4}+\frac{3}{4}}+a^{\frac{1}{4}-\frac{1}{4}}}$
= $\frac{a+a^{2}}{a+1}=\frac{a(a+1)}{a+1}=a$
Vậy $\frac{a^{\frac{4}{3}}(a^{-\frac{1}{3}}+a^{\frac{2}{3}})}{a^{\frac{1}{4}}(a^{\frac{3}{4}}+a^{-\frac{1}{4}})}=a$
b) $\frac{b^{\frac{1}{5}}(\sqrt[5]{b^{4}}-\sqrt[5]{b^{-1}})}{b^{\frac{2}{3}}(\sqrt[3]{b}-\sqrt[3]{b^{-2}})}$
= $\frac{b^{\frac{1}{5}}.b^{\frac{1}{4}}-b^{\frac{1}{5}}.b^{-\frac{1}{5}}}{b^{\frac{2}{3}}.b^{\frac{1}{3}}-b^{\frac{2}{3}}.b^{-\frac{2}{3}}}$
= $\frac{b-1}{b-1}=1$
Vậy $\frac{b^{\frac{1}{5}}(\sqrt[5]{b^{4}}-\sqrt[5]{b^{-1}})}{b^{\frac{2}{3}}(\sqrt[3]{b}-\sqrt[3]{b^{-2}})}=1$
c) $\frac{a^{\frac{1}{3}}b^{-\frac{1}{3}}-a^{-\frac{1}{3}}b^{\frac{1}{3}}}{\sqrt[3]{a^{2}}-\sqrt[3]{b^{2}}}$
= $\frac{a^{\frac{2}{3}-\frac{1}{3}}b^{\frac{-1}{3}}-a^{\frac{-1}{2}}b^{\frac{2}{3}-\frac{1}{3}}}{a^{\frac{1}{6}}+a^{\frac{1}{6}}}$
= $(ab)^{\frac{-1}{3}}=\frac{1}{\sqrt[3]{ab}}$
Vậy $\frac{a^{\frac{1}{3}}b^{-\frac{1}{3}}-a^{-\frac{1}{3}}b^{\frac{1}{3}}}{\sqrt[3]{a^{2}}-\sqrt[3]{b^{2}}}=\frac{1}{\sqrt[3]{ab}}$
d) $\frac{a^{\frac{1}{3}}\sqrt{b}+b^{\frac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}$
= $\frac{a^{\frac{1}{3}}.b^{\frac{1}{2}}+b^{\frac{1}{3}}.a^{\frac{1}{2}}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}$
= $\frac{(ab)^{\frac{1}{3}}\left [ b^{\frac{1}{6}}+a^{\frac{1}{6}} \right ]}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}=\sqrt[3]{ab}$
Vậy $\frac{a^{\frac{1}{3}}\sqrt{b}+b^{\frac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}=\sqrt[3]{ab}$
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Câu 5: Trang 56- sgk giải tích 12
Chứng minh rằng:
a) $(\frac{1}{3})^{2\sqrt{5}}<(\frac{1}{3})^{3\sqrt{2}}$
b) $7^{6\sqrt{3}}>7^{3\sqrt{6}}$
Hướng dẫn giải:
Ta có:
a) $2\sqrt{5}=\sqrt{2^{2}.5}=\sqrt{20}$
$3\sqrt{2}=\sqrt{3^{2}.2}=\sqrt{18}$
Mà $\sqrt{18}<\sqrt{20}$
=> $2\sqrt{5}>3\sqrt{2}$ và $0<\frac{1}{3}<1$
=> $(\frac{1}{3})^{2\sqrt{5}}<(\frac{1}{3})^{3\sqrt{2}}$. ( đpcm)
b) $6\sqrt{3}=\sqrt{6^{2}.3}=\sqrt{108}$
$3\sqrt{6}=\sqrt{3^{2}.6}=\sqrt{54}$
Mà $\sqrt{54}<\sqrt{108}$
=> $3\sqrt{6}<6\sqrt{3}$
=> $7^{6\sqrt{3}}>7^{3\sqrt{6}}$. ( đpcm)
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