Biết \({4^\alpha } + {4^{ - \alpha }} = 5\).
Tính giá trị của các biểu thức:
a) \({2^\alpha } + {2^{ - \alpha }}\);
b) \({4^{2\alpha }} + {4^{ - 2\alpha }}\).
K
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Những câu hỏi liên quan
H
18 tháng 8 2023
a)
$16^{\alpha }+16^{-\alpha } = (4^2)^{\alpha }+(4^2)^{-\alpha } = 4^{2\alpha }+4^{-2\alpha }$
$4^{2\alpha }+4^{-2\alpha } = 4^{2\log_4{\frac{1}{5}}}+4^{-2\log_4{\frac{1}{5}}} = \left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^{-2} = \frac{1}{25}+25 = \frac{26}{25}$
H
18 tháng 8 2023
b)
$\left(2^{\alpha }+2^{-\alpha }\right)^2 = \left(\sqrt{4}\right)^{\alpha }+\left(\sqrt{4}\right)^{-\alpha } = 4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}}$
$4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}} = 4^{\frac{\log_4{\frac{1}{5}}}{2}}+4^{-\frac{\log_4{\frac{1}{5}}}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}}+\left(\frac{1}{5}\right)^{-\frac{1}{2}} = \sqrt{\frac{1}{5}}+\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2021
\(A=\dfrac{\dfrac{3sina}{cosa}-\dfrac{5cosa}{cosa}}{\dfrac{5sina}{cosa}+\dfrac{8cosa}{cosa}}=\dfrac{3tana-5}{5tana+8}=\dfrac{3.\left(\dfrac{5}{7}\right)-5}{5.\left(\dfrac{5}{7}\right)+8}=...\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2023
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(a,cos2\alpha=2cos^2\alpha-1=\dfrac{2}{5}\\ \Leftrightarrow cos^2\alpha=\dfrac{7}{10}\Rightarrow cos\alpha=\pm\dfrac{\sqrt{70}}{10}\)
Vì \(-\dfrac{\pi}{2}< \alpha< 0\Rightarrow cos\alpha=\dfrac{\sqrt{70}}{10}\)
Ta có:
\(sin^2\alpha+cos^2\alpha=1\\ \Rightarrow sin^2\alpha=1-\dfrac{7}{10}=\dfrac{3}{10}\\ \Rightarrow sin\alpha=\pm\sqrt{30}10\)
Vì \(-\dfrac{\pi}{2}< \alpha< 0\Rightarrow sin\alpha=-\dfrac{\sqrt{30}}{10}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\dfrac{\sqrt{30}}{10}}{\dfrac{-\sqrt{70}}{10}}=-\dfrac{\sqrt{21}}{7}\\ cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{-\dfrac{\sqrt{21}}{7}}=-\dfrac{\sqrt{21}}{3}\)
HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(b,sin^22\alpha+cos^22\alpha=1\\ \Rightarrow cos2\alpha=\sqrt{1-\left(-\dfrac{4}{9}\right)^2}=\pm\dfrac{\sqrt{65}}{9}\)
Vì \(\dfrac{\pi}{2}< \alpha< \dfrac{3\pi}{4}\Rightarrow\pi< 2\alpha< \dfrac{3\pi}{2}\Rightarrow cos2\alpha=-\dfrac{\sqrt{65}}{9}\)
\(cos2\alpha=2cos^2\alpha-1=-\dfrac{\sqrt{65}}{9}\\ \Rightarrow cos\alpha=\pm\sqrt{\dfrac{9-\sqrt{65}}{18}}\)
Vì \(\dfrac{\pi}{2}< \alpha< \dfrac{3\pi}{4}\Rightarrow cos\alpha=-\sqrt{\dfrac{9-\sqrt{65}}{18}}\)
\(sin^2\alpha+cos^2\alpha=1\\ \Rightarrow sin^2\alpha=\dfrac{9+\sqrt{65}}{18}\\ \Rightarrow sin\alpha=\pm\sqrt{\dfrac{9+\sqrt{65}}{18}}\)
Vì \(\dfrac{\pi}{2}< \alpha< \dfrac{3\pi}{4}\Rightarrow sin\alpha=\sqrt{\dfrac{9+\sqrt{65}}{18}}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{\dfrac{9+\sqrt{65}}{18}}}{-\sqrt{\dfrac{9-\sqrt{65}}{18}}}\approx-4,266\\ cot\alpha=\dfrac{1}{tan\alpha}\approx-0,234\)
BU
7 tháng 4 2022
mình làm r nha
https://hoc24.vn/cau-hoi/biet-cotadfrac12-gia-tri-bieu-thuc-adfrac4sinalpha5cosalpha2sinalpha-3cosalpha-bang-bao-nhieughi-ro-tung-loi-giai-nha.5724337531039
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a)
Ta có:
\({\cos ^4}\alpha {\sin ^4}\alpha = \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) \\= {\cos ^2}\alpha - {\sin ^2}\alpha = {\cos ^2}\alpha - (1 - {\cos ^2}\alpha ) \\= {\cos ^2}\alpha - 1 + {\cos ^2}\alpha = 2{\cos ^2}\alpha - 1\)
(đpcm)
b)
Ta có:
\(\frac{{{{\cos }^2}\alpha + {{\tan }^2}\alpha - 1}}{{{{\sin }^2}\alpha }} = \frac{{{{\cos }^2}\alpha \; + {{\tan }^2}\alpha - {{\sin }^2}\alpha - {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{1}{{{{\cos }^2}\alpha }} - 1 = {\tan ^2}\alpha \)
(đpcm)
AT
29 tháng 7 2021
Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)
Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)
\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
29 tháng 7 2021
Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)
\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)
\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)
\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
13 tháng 8 2021
Bài 1:
Ta có: \(A=\sin^6\alpha+3\cdot\sin^2\alpha\cdot\cos^2\alpha+\cos^6\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3-3\cdot\sin^2\alpha\cdot\cos\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot\cos^2\alpha\)
\(=1^3\)
=1
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2022
\(\dfrac{2sina+cosa}{2sin^3a-cos^3a}=\dfrac{\dfrac{2sina}{cos^3a}+\dfrac{cosa}{cos^3a}}{\dfrac{2sin^3a}{cos^3a}-\dfrac{cos^3a}{cos^3a}}=\dfrac{2tana.\dfrac{1}{cos^2a}+\dfrac{1}{cos^2a}}{2tan^3a-1}\)
\(=\dfrac{2tana\left(1+tan^2a\right)+1+tan^2a}{2tan^3a-1}=...\) (thay số và bấm máy)
27 tháng 3 2022
Em vẫn ch hiểu tại sao cosa/cos3a lại ra 1/cos2a thầy giải thích giúp em vs ạ
a, Ta có:
\(\left(2^{\alpha}+2^{-\alpha}\right)^2\\ =\left(2^{\alpha}\right)^2+2\cdot2^{\alpha}\cdot2^{-\alpha}+\left(2^{-\alpha}\right)^2\\ =4^{\alpha}+4^{-\alpha}+2\\ =5+2\\ =7\)
Vậy \(2^{\alpha}+2^{-\alpha}=7\)
b, Ta có:
\(4^{2\alpha}+4^{-2\alpha}=\left(4^{\alpha}\right)^2+\left(4^{-\alpha}\right)^2\\ =\left(4^{\alpha}+4^{-\alpha}\right)^2-2\\ =5^2-2\\ =23\)
Vậy \(4^{2\alpha}+4^{-2\alpha}=23\)