Bài 2.1 trang 99 SBT giải tích 12

Đề bài

Tính:

a)  \( \dfrac{10^{2+ \sqrt{7}}}{2^{2 + \sqrt{7}}. 5^{1+\sqrt{7}}}\)

b) \( ( 4^{2\sqrt{3}} – 4^{\sqrt{3} – 1}). 2^{-2\sqrt{3}}.\)

Lời giải chi tiết

a)  \( \dfrac{10^{2+ \sqrt{7}}}{2^{2 + \sqrt{7}}. 5^{1+\sqrt{7}}}\)

\( = \dfrac{ (2.5)^{2+ \sqrt{7}}}{2^{2 + \sqrt{7}}. 5^{1+\sqrt{7}}}\)

\(= \dfrac{2^{2+ \sqrt{7}}. 5^{2+ \sqrt{7}}}{2^{2 + \sqrt{7}}. 5^{1+\sqrt{7}}}\)

\( = \dfrac{5^{2+ \sqrt{7}}}{5^{1+\sqrt{7}}} \)

\( = 5^{(2+ \sqrt{7}) – ( 1+ \sqrt{7})} \)

\( = 5^1 =5 \).

b) \( ( 4^{2\sqrt{3}} – 4^{\sqrt{3} – 1}). 2^{-2\sqrt{3}}\)

\( = 4^{2\sqrt{3}}.2^{-2\sqrt{3}} – 4^{\sqrt{3} – 1}.2^{-2\sqrt{3}}\)

\(= \Big(2^{2}\Big)^{2\sqrt{3}}.2^{-2\sqrt{3}} – \Big(2^{2}\Big)^{\sqrt{3} – 1}. 2^{-2\sqrt{3}}\)

\(= 2^{4\sqrt{3}}.2^{-2\sqrt{3}} – 2^{2\sqrt{3} – 2}. 2^{-2\sqrt{3}}\)

\(= 2^{2\sqrt{3}} – 2^ {-2}\)

\(= 2^{2\sqrt{3}} – \dfrac{1}{4}\)

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