Tính: 9^2/5.27^2/5

Giải Toán 12 Bài 1: Lũy thừa

Bài 1 trang 55 Toán lớp 12 Giải tích: Tính:

a) $9^{\frac{2}{5}}{.27}^{\frac{2}{5}}$;

b) ${144}^{\frac{3}{4}}:9^{\frac{3}{4}}$;

c) ${\left(\frac{1}{16}\right)}^{-0,75}+{\left(0,25\right)}^{-\frac{5}{2}}$;

d) ${\left(0,04\right)}^{-1,5}-{\left(0,125\right)}^{-\frac{2}{3}}$.

Lời giải:

a) $9^{\frac{2}{5}}{.27}^{\frac{2}{5}}={\left(9.27\right)}^{\frac{2}{5}}$

$={\left(3^2{.3}^3\right)}^{\frac{2}{5}}={\left(3^5\right)}^{\frac{2}{5}}=3^{5.\frac{2}{5}}=3^2=9.$

b) ${144}^{\frac{3}{4}}:9^{\frac{3}{4}}={\left(144:9\right)}^{\frac{3}{4}}$

$={16}^{\frac{3}{4}}={\left(2^4\right)}^{\frac{3}{4}}=2^{4.\frac{3}{4}}=2^3=8$

 $\mathbf{c}\mathbf{)}{\left(\frac{1}{16}\right)}^{-0,75}+{\left(0,25\right)}^{-\frac{5}{2}}={\left(2^{-4}\right)}^{\frac{-3}{4}}+{\left(2^{-2}\right)}^{-\frac{5}{2}}=2^{-4.\frac{-3}{4}}+2^{-2.\frac{-5}{2}}=2^3+2^5=8+32=40$

= 125 – 4 = 121.

d) ${\left(0,04\right)}^{-1,5}-{\left(0,125\right)}^{-\frac{2}{3}}$

$={\left(\frac{1}{25}\right)}^{\frac{-3}{2}}-{\left(\frac{1}{8}\right)}^{\frac{-2}{3}}$

$={\left(5^{-2}\right)}^{\frac{-3}{2}}-{\left(2^{-3}\right)}^{\frac{-2}{3}}=5^{-2.\frac{-3}{2}}-2^{-3.\frac{-2}{3}}=5^3-2^2=125–4=121.$