Bài 1 trang 33 Toán 11 Tập 2 | Cánh diều Giải Toán lớp 11

Giải Toán 11 Bài 1: Phép tính lũy thừa với số mũ thực

Bài 1 trang 33 Toán 11 Tập 2: Tính:

a) ${\left(\frac{1}{256}\right)}^{-\mathrm{0,75}}+{\left(\frac{1}{27}\right)}^{-\frac{4}{3}};$

b) ${\left(\frac{1}{49}\right)}^{-\mathrm{1,5}}-{\left(\frac{1}{125}\right)}^{-\frac{2}{3}};$

c) $\left(4^{3+\sqrt{3}}-4^{\sqrt{3}-1}\right)\cdot 2^{-2\sqrt{3}}.$

Lời giải:

Ta có:

a) ${\left(\frac{1}{256}\right)}^{-\mathrm{0,75}}+{\left(\frac{1}{27}\right)}^{-\frac{4}{3}}={\left(\frac{1}{256}\right)}^{-\frac{3}{4}}+{\left(\frac{1}{27}\right)}^{-\frac{4}{3}}$

$={256}^{\frac{3}{4}}+{27}^{\frac{4}{3}}=\sqrt[4]{{256}^3}+\sqrt[3]{{27}^4}$

$=\sqrt[4]{{\left(2^8\right)}^3}+\sqrt[3]{{\left(3^3\right)}^4}=\sqrt[4]{2^{24}}+\sqrt[3]{3^{12}}$

$=\sqrt[4]{{\left(2^6\right)}^4}+\sqrt[3]{{\left(3^4\right)}^3}=2^6+3^4=64+81=145.$

b) ${\left(\frac{1}{49}\right)}^{-\mathrm{1,5}}-{\left(\frac{1}{256}\right)}^{-\frac{2}{3}}={\left(\frac{1}{49}\right)}^{-\frac{3}{2}}-{\left(\frac{1}{256}\right)}^{-\frac{2}{3}}$

$={49}^{\frac{3}{2}}-{256}^{\frac{2}{3}}=\sqrt{{49}^3}-\sqrt[3]{{256}^2}$

$=\sqrt{{\left(7^2\right)}^3}-\sqrt[3]{{\left(2^8\right)}^2}=\sqrt{7^6}-\sqrt[3]{2^{16}}$

$=\sqrt{{\left(7^3\right)}^2}-\sqrt[3]{2\cdot 2^{15}}=7^3-\sqrt[3]{2}\cdot \sqrt[3]{{\left(2^5\right)}^3}$

$=7^3-\sqrt[3]{2}\cdot 2^5=7^3-32\sqrt[3]{2}.$

c)

$=\left(2^{6+2\sqrt{3}}-2^{2\sqrt{3}-2}\right)\cdot 2^{-2\sqrt{3}}=2^{6+2\sqrt{3}-2\sqrt{3}}-2^{2\sqrt{3}-2-2\sqrt{3}}$

$=2^6-2^{-2}=64-\frac{1}{2^2}=64-\frac{1}{4}=\frac{255}{4}.$