Bài 7 trang 13

Đề bài

Biết rằng $4^{\alpha }=\frac{1}{5}$. Tính giá trị các biểu thức sau:

a) ${16}^{\alpha }+{16}^{-\alpha }$;                 

b) ${\left(2^{\alpha }+2^{-\alpha }\right)}^2$.

Lời giải chi tiết

a) ${16}^{\alpha }+{16}^{-\alpha }$

$={16}^{\alpha }+\frac{1}{{16}^{\alpha }}$

$={\left(4^2\right)}^{\alpha }+\frac{1}{{\left(4^2\right)}^{\alpha }}$

$={\left(4^{\alpha }\right)}^2+\frac{1}{{\left(4^{\alpha }\right)}^2}$

$={\left(\frac{1}{5}\right)}^2+\frac{1}{{\left(\frac{1}{5}\right)}^2}$

$=\frac{626}{25}$.

b) ${\left(2^{\alpha }+2^{-\alpha }\right)}^2$

$={\left(2^{\alpha }\right)}^2+{2.2}^{\alpha }{.2}^{-\alpha }+{\left(2^{-\alpha }\right)}^2$

$=2^{2\alpha }+2+2^{-2\alpha }$

$={\left(2^2\right)}^{\alpha }+2+{\left(2^2\right)}^{-\alpha }$

$=4^{\alpha }+2+4^{-\alpha }$

$=4^{\alpha }+2+\frac{1}{4^{\alpha }}$

$=\frac{1}{5}+2+\frac{1}{\frac{1}{5}}$

$=\frac{36}{5}$.