cho α ∈(0; $\frac{π}{2}$ ).Biểu thức $2^{sin^{4}α}$2$^{cos^{2}α}$.4$^{sin^{2}αcos^{2}α}$ bằng
1 câu trả lời
$\begin{array}{l}2^{\sin^4\alpha}.2^{\cos^2\alpha}.4^{\sin^2\alpha.\cos^2\alpha}\\ = 2^{\sin^4\alpha + \cos^2\alpha + 2\sin^2\alpha\cos^2\alpha}\\ Xét \,\, \sin^4\alpha + \cos^2\alpha + 2\sin^2\alpha\cos^2\alpha\\ = \sin^2(\sin^2\alpha +2\cos^2\alpha) + \cos^2\alpha\\ = \sin^2\alpha(1 + \cos^2\alpha) + \cos^2\alpha\\ = 1 + \sin^2\alpha\cos^2\alpha\\ Ta\,\,được:\\ 2^{\sin^4\alpha + \cos^2\alpha + 2\sin^2\alpha\cos^2\alpha}= 2^{1 + \sin^2\alpha\cos^2\alpha}\end{array}$
Câu hỏi trong lớp
Xem thêm
