Bài tự luận · Bài 22
Dạng 2. Rút gọn biểu thức chứa luỹ thừa
Bài tự luận Lũy thừa với số mũ thực · Bài 22
(CTST11) Rút gọn các biểu thức sau:
a) $ 2^{\sqrt{3}+1}:2^{\sqrt{3}-1} $
b) $ \left( 3^{\sqrt{2}} \right)^{\sqrt{8}} $
c) $ \left[ (\sqrt{7})^{\sqrt{2}} \right]^{\sqrt{8}} $
d) $ a^{2\sqrt{5}+1}:a^{2\sqrt{5}-2} $
e) $ 3^{3+\sqrt{2}}\cdot 3^{-1+\sqrt{2}}\cdot 9^{1-\sqrt{2}} $
g) $ \left( a^{-\sqrt{3}}b^{\frac{1}{\sqrt{3}}} \right)^{\frac{1}{\sqrt{3}}} $.
a) $ 2^{\sqrt{3}+1}:2^{\sqrt{3}-1} $
b) $ \left( 3^{\sqrt{2}} \right)^{\sqrt{8}} $
c) $ \left[ (\sqrt{7})^{\sqrt{2}} \right]^{\sqrt{8}} $
d) $ a^{2\sqrt{5}+1}:a^{2\sqrt{5}-2} $
e) $ 3^{3+\sqrt{2}}\cdot 3^{-1+\sqrt{2}}\cdot 9^{1-\sqrt{2}} $
g) $ \left( a^{-\sqrt{3}}b^{\frac{1}{\sqrt{3}}} \right)^{\frac{1}{\sqrt{3}}} $.
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Lời giải
a) $ 2^{\sqrt{3}+1}:2^{\sqrt{3}-1}=2^{\sqrt{3}+1-(\sqrt{3}-1)}=2^{2}=4 $;
b) $ \left( 3^{\sqrt{2}} \right)^{\sqrt{8}}=3^{\sqrt{2}\cdot \sqrt{8}}=3^{\sqrt{16}}=3^{4}=81 $;
c) $ \left[ (\sqrt{7})^{\sqrt{2}} \right]^{\sqrt{8}}=\left( 7^{\frac{1}{2}} \right)^{\sqrt{16}}=7^{\frac{1}{2}\cdot 4}=7^{2}=49 $;
d) $ a^{2\sqrt{5}+1}:a^{2\sqrt{5}-2}=a^{2\sqrt{5}+1-(2\sqrt{5}-2)}=a^{3} $;
e) $ 3^{3+\sqrt{2}}\cdot 3^{-1+\sqrt{2}}\cdot 9^{1-\sqrt{2}}=3^{3+\sqrt{2}-1+\sqrt{2}}\cdot \left( 3^{2} \right)^{1-\sqrt{2}}=3^{2+2\sqrt{2}}\cdot 3^{2-2\sqrt{2}}=3^{4}=81 $;
g) $ \left( a^{-\sqrt{3}}b^{\frac{1}{\sqrt{3}}} \right)^{\frac{1}{\sqrt{3}}}=a^{-\sqrt{3}\cdot \frac{1}{\sqrt{3}}}\cdot b^{\frac{1}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}}}=a^{-1}b^{\frac{1}{3}}=\frac{\sqrt[3]{b}}{a} $.
b) $ \left( 3^{\sqrt{2}} \right)^{\sqrt{8}}=3^{\sqrt{2}\cdot \sqrt{8}}=3^{\sqrt{16}}=3^{4}=81 $;
c) $ \left[ (\sqrt{7})^{\sqrt{2}} \right]^{\sqrt{8}}=\left( 7^{\frac{1}{2}} \right)^{\sqrt{16}}=7^{\frac{1}{2}\cdot 4}=7^{2}=49 $;
d) $ a^{2\sqrt{5}+1}:a^{2\sqrt{5}-2}=a^{2\sqrt{5}+1-(2\sqrt{5}-2)}=a^{3} $;
e) $ 3^{3+\sqrt{2}}\cdot 3^{-1+\sqrt{2}}\cdot 9^{1-\sqrt{2}}=3^{3+\sqrt{2}-1+\sqrt{2}}\cdot \left( 3^{2} \right)^{1-\sqrt{2}}=3^{2+2\sqrt{2}}\cdot 3^{2-2\sqrt{2}}=3^{4}=81 $;
g) $ \left( a^{-\sqrt{3}}b^{\frac{1}{\sqrt{3}}} \right)^{\frac{1}{\sqrt{3}}}=a^{-\sqrt{3}\cdot \frac{1}{\sqrt{3}}}\cdot b^{\frac{1}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}}}=a^{-1}b^{\frac{1}{3}}=\frac{\sqrt[3]{b}}{a} $.