Tính giá trị của \(\alpha=log_{\alpha}\left(\frac{a^2\sqrt[3]{a^2}.a\sqrt[5]{a^4}}{\sqrt[3]{a}}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
AT
30 tháng 7 2021
\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)
\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)
6 tháng 5 2021
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)
