Rút gọn các biểu thức sau: 2^(căn 3 +1) : 2^(căn 3 -1)

Giải SBT Toán 11 Bài 1: Phép tính lũy thừa

Bài 8 trang 9 SBT Toán 11 Tập 2: Rút gọn các biểu thức sau:

a) $2^{\sqrt{3}+1}:2^{\sqrt{3}-1}$;

b) ${\left(3^{\sqrt{2}}\right)}^{\sqrt{8}}$;

c) ${\left[{\left(\sqrt{7}\right)}^{\sqrt{2}}\right]}^{\sqrt{8}}$;

d) $a^{2\sqrt{5}+1}:a^{2\sqrt{5}-2}$;

e) $3^{3+\sqrt{2}}.3^{-1+\sqrt{2}}.9^{1-\sqrt{2}}$;

g) ${\left(a^{-\sqrt{3}}{b}^{\frac{1}{\sqrt{3}}}\right)}^{\frac{1}{\sqrt{3}}}$.

Lời giải:

a) $2^{\sqrt{3}+1}:2^{\sqrt{3}-1}=2^{\sqrt{3}+1-(\sqrt{3}-1)}=2^2=4$.

b) ${\left(3^{\sqrt{2}}\right)}^{\sqrt{8}}=3^{\sqrt{2}.\sqrt{8}}=3^{\sqrt{2.8}}=3^{\sqrt{16}}=3^4=81$.

c) ${\left[{\left(\sqrt{7}\right)}^{\sqrt{2}}\right]}^{\sqrt{8}}={\left(\sqrt{7}\right)}^{\sqrt{2}.\sqrt{8}}={\left(\sqrt{7}\right)}^{\sqrt{2.8}}$

= ${\left(\sqrt{7}\right)}^{\sqrt{16}}={\left(\sqrt{7}\right)}^4=7^2=49$.

d) $a^{2\sqrt{5}+1}:a^{2\sqrt{5}-2}=a^{2\sqrt{5}+1-\left(2\sqrt{5}-2\right)}=a^3$.

e) $3^{3+\sqrt{2}}.3^{-1+\sqrt{2}}.9^{1-\sqrt{2}}=3^{3+\sqrt{2}}.3^{-1+\sqrt{2}}.{\left(3^2\right)}^{1-\sqrt{2}}$

= $3^{3+\sqrt{2}+\left(-1+\sqrt{2}\right)+2.\left(1-\sqrt{2}\right)}$ = 3⁴ = 81.

g) ${\left(a^{-\sqrt{3}}{b}^{\frac{1}{\sqrt{3}}}\right)}^{\frac{1}{\sqrt{3}}}={\left(a^{-\sqrt{3}}\right)}^{\frac{1}{\sqrt{3}}}.{\left(b^{\frac{1}{\sqrt{3}}}\right)}^{\frac{1}{\sqrt{3}}}$

= $a^{-\sqrt{3}.\frac{1}{\sqrt{3}}}.b^{\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}}=a^{-1}.b^{\frac{1}{3}}=\frac{\sqrt[3]{b}}{a}$.