Viết các số sau theo thứ tự tăng dần
a) \(0,3^{\pi};0,3^{0.5};0,3^{\frac{2}{3}};0,3^{3,15}\)
b) \(\sqrt{2\pi};1,8^{\pi};\left(\frac{1}{\sqrt{2}}\right)^{\pi};\pi^{\pi}\)
Hãy viết các số sau theo thứ tự tăng dần :
a) \(\left(0,3\right)^{\pi};\left(0,3\right)^{0,5};\left(0,3\right)^{\dfrac{2}{3}};\left(0,3\right)^{3,1415}\)
b) \(\sqrt{2^{\pi}};\left(1,9\right)^{\pi};\left(\dfrac{1}{\sqrt{2}}\right)^{\pi};\pi^{\pi}\)
c) \(5^{-2};5^{-0,7};5^{\dfrac{1}{3}};\left(\dfrac{1}{5}\right)^{2,1}\)
d) \(\left(0,5\right)^{-\dfrac{2}{3}};\left(1,3\right)^{-\dfrac{2}{3}};\pi^{-\dfrac{2}{3}};\left(\sqrt{2}\right)^{-\dfrac{2}{3}}\)
Cho \(\alpha ,\beta \) là hai số thực với \(\alpha < \beta \). Khẳng định nào sau đây đúng?
A. \({\left( {0,3} \right)^\alpha } < {\left( {0,3} \right)^\beta }\).
B. \({\pi ^\alpha } \ge {\pi ^\beta }\).
C. \({\left( {\sqrt 2 } \right)^\alpha } < {\left( {\sqrt 2 } \right)^\beta }\).
D. \({\left( {\frac{1}{2}} \right)^\beta } > {\left( {\frac{1}{2}} \right)^\alpha }\).
Ta có:
A. \(\alpha< \beta\)
\(\Rightarrow\left(0,3\right)^{\alpha}>\left(0,3\right)^{\beta}\)
Sai
B. \(\alpha< \beta\)
\(\Rightarrow\pi^{\alpha}< \pi^{\beta}\)
Sai
C. \(\alpha< \beta\)
\(\Rightarrow\left(\sqrt{2}\right)^{\alpha}< \left(\sqrt{2}\right)^{\beta}\)
Đúng
D. \(\alpha< \beta\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{\alpha}>\left(\dfrac{1}{2}\right)^{\beta}\)
Sai
⇒ Chọn C
Tìm x \(\in\left(0,3\pi\right)\)sao cho : sin \(\left(x-\frac{\pi}{3}\right)\)+ 2cos\(\left(x+\frac{\pi}{6}\right)\)=0
So sánh các cặp số sau:
a) \({\log _\pi }0,8\) và \({\log _\pi }1,2\);
b) \({\log _{0,3}}2\) và \({\log _{0,3}}2,1\);
a) Vì \(\pi>1\) nên hàm số \(log_{\pi}x\) đồng biến trên\(\left(0;+\infty\right)\)
Mà \(0,8< 1,2\) nên \(log_{\pi}0,8< log_{\pi}1,2\)
b) Vì \(0,3>1\) nên hàm số \(log_{0,3}x\) nghịch biến trên \(\left(0;+\infty\right)\)
Mà \(2<2,1\) nên \(log_{0,3}2>log_{0,3}2,1\)Tính:
a) \(\cos \left( {a + \frac{\pi }{6}} \right)\), biết \(\sin a = \frac{1}{{\sqrt 3 }}\) và \(\frac{\pi }{2} < a < \pi \);
b) \(\tan \left( {a - \frac{\pi }{4}} \right)\), biết \(\cos a = - \frac{1}{3}\) và \(\pi < a < \frac{{3\pi }}{2}\).
a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\). Do đó \(\cos a = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \frac{1}{3}} = - \frac{{\sqrt 6 }}{3}\)
Ta có: \(\cos \left( {a + \frac{\pi }{6}} \right) = \cos a\cos \frac{\pi }{6} - \sin a\sin \frac{\pi }{6} = - \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 3 }}.\frac{1}{2} = - \frac{{\sqrt 3 + 3\sqrt 2 }}{6}\)
b) Vì \(\pi < a < \frac{{3\pi }}{2}\) nên \(\sin a < 0\). Do đó \(\sin a = \sqrt {1 - {{\cos }^2}a} = \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a\; = \frac{{\sin a}}{{\cos a}} = \frac{{ - \frac{{2\sqrt 2 }}{3}}}{{ - \frac{1}{3}}} = 2\sqrt 2 \)
Ta có: \(\tan \left( {a - \frac{\pi }{4}} \right) = \frac{{\tan a - \tan \frac{\pi }{4}}}{{1 + \tan a\tan \frac{\pi }{4}}} = \frac{{\frac{{\sin a}}{{\cos a}} - 1}}{{1 + \frac{{\sin a}}{{\cos a}}}} = \frac{{2\sqrt 2 - 1}}{{1 + 2\sqrt 2 }} = \frac{{9 - 4\sqrt 2 }}{7}\)
Chứng minh rằng:
a) \(\sin x - \cos x = \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right)\);
b) \(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{1 - \tan x}}{{1 + \tan x}}\;\left( {x \ne \frac{\pi }{2} + k\pi ,\;x \ne \frac{{3\pi }}{4} + k\pi ,\;k \in \mathbb{Z}} \right)\;\).
a) Ta có:
\(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)
b) Ta có:
\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)
Giải phương trình:
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
d) \(2\cos 3x + 5 = 3\)
e) \(3\tan x = - \sqrt 3 \)
g) \(\cot x - 3 = \sqrt 3 \left( {1 - \cot x} \right)\)
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(2\cos 3x + 5 = 3\)
\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
e)
\(\begin{array}{l}3\tan x = - \sqrt 3 \\ \Leftrightarrow \tan x = \frac{{ - \sqrt 3 }}{3}\\ \Leftrightarrow \tan x = \tan \left( { - \frac{\pi }{6}} \right)\\ \Leftrightarrow x = - \frac{\pi }{6} + k\pi \end{array}\)
g)
\(\begin{array}{l}\cot x - 3 = \sqrt 3 \left( {1 - \cot x} \right)\\ \Leftrightarrow \cot x - 3 = \sqrt 3 - \sqrt 3 \cot x\\ \Leftrightarrow \cot x + \sqrt 3 \cot x = \sqrt 3 + 3\\ \Leftrightarrow (1 + \sqrt 3 )\cot x = \sqrt 3 + 3\\ \Leftrightarrow \cot x = \sqrt 3 \\ \Leftrightarrow \cot x = \cot \frac{\pi }{6}\\ \Leftrightarrow x = \frac{\pi }{6} + k\pi \end{array}\)
Cho góc \(\alpha \) thỏa mãn \(\frac{\pi }{2} < \alpha < \pi ,\cos \alpha = - \frac{1}{{\sqrt 3 }}\). Tính giá trị của các biểu thức sau:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right)\);
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right);\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right)\);
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right)\).
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
\(2\sin^2\left(5\pi+1\right)-\left(\sqrt{3}+1\right)\sin2\left(\frac{\pi}{2}-x\right)+\sqrt{3}\sin^2\left(\frac{9\pi}{2}+x\right)=0\)
Đề đúng là \(2sin^2\left(5\pi+1\right)\) chứ bạn?
Chứ thấy nó hơi thế nào ấy?