Bài 1 trang 33

Đề bài

Tính:

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

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

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

Lời giải chi tiết

a)

${\left(\frac{1}{256}\right)}^{-0,75}+{\left(\frac{1}{27}\right)}^{-\frac{4}{3}}={\left(\frac{1}{256}\right)}^{-\frac{3}{4}}+{\left(\frac{1}{27}\right)}^{-\frac{4}{3}}={256}^{\frac{3}{4}}+{27}^{\frac{4}{3}}=\sqrt[4]{4{256}^3}+\sqrt[3]{3{27}^4}=\sqrt[4]{4{\left(2^8\right)}^3}+\sqrt[3]{3{\left(3^3\right)}^4}$

$=\sqrt[4]{42^{24}}+\sqrt[3]{33^{12}}=2^6+3^4=145$

b)    

${\left(\frac{1}{49}\right)}^{-1,5}-{\left(\frac{1}{125}\right)}^{-\frac{2}{3}}={\left(\frac{1}{49}\right)}^{\frac{3}{2}}-{\left(\frac{1}{125}\right)}^{\frac{2}{3}}={\left(\frac{1}{7}\right)}^{2.\frac{-3}{2}}-{\left(\frac{1}{5}\right)}^{3.\frac{-2}{3}}$

$={\left(7^{-1}\right)}^{-3}-{\left(5^{-1}\right)}^{-2}=7^3-5^2=318$

c)

$\left(4^{3+\sqrt{3}}-4^{\sqrt{3}-1}\right){.2}^{-2\sqrt{3}}=\left(2^{6+2\sqrt{3}}-2^{2\sqrt{3}-2}\right){.2}^{-2\sqrt{3}}=2^6-2^{-2}=64-\frac{1}{4}=\frac{255}{4}$