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y = \(\dfrac{sin^2x}{cosx\left(sinx-cosx\right)}+\dfrac{1}{4}\)
y = \(\dfrac{sin^2x}{sinx.cosx-cos^2x}+\dfrac{1}{4}=\dfrac{\dfrac{sin^2x}{cos^2x}}{\dfrac{sinx.cosx}{cos^2x}-1}+\dfrac{1}{4}\)
y = \(\dfrac{tan^2x}{tanx-1}+\dfrac{1}{4}\)
y = \(\dfrac{4tan^2x+tanx-1}{4tanx-4}\). Đặt t = tanx. Do x ∈ \(\left(\dfrac{\pi}{4};\dfrac{\pi}{2}\right)\) nên t ∈ (1 ; +\(\infty\))\
Ta đươc hàm số f(t) = \(\dfrac{4t^2+t-1}{4t-4}\)
⇒ ymin = \(\dfrac{17}{4}\) khi t = 2. hay x = arctan(2) + kπ
a: pi/2<a<pi
=>sin a>0
\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)
\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)
b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c: \(sin\left(a-\dfrac{pi}{3}\right)\)
\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)
d: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
Đặt \(sinx=t\Rightarrow t\in\left[-\dfrac{1}{2};1\right]\)
\(y=f\left(t\right)=2t^2+t+4\)
Xét hàm \(f\left(t\right)=2t^2+t+4\) trên \(\left[-\dfrac{1}{2};1\right]\)
\(-\dfrac{b}{2a}=-\dfrac{1}{4}\in\left[-\dfrac{1}{2};1\right]\)
\(f\left(-\dfrac{1}{2}\right)=4\) ; \(f\left(-\dfrac{1}{4}\right)=\dfrac{31}{8}\); \(f\left(1\right)=7\)
\(y_{max}=7\) khi \(t=1\) hay \(x=\dfrac{\pi}{2}\)
\(y_{min}=\dfrac{31}{8}\) khi \(sinx=-\dfrac{1}{4}\)
ĐKXĐ:
a. \(cos\left(x-\dfrac{2\pi}{3}\right)\ne0\Rightarrow x-\dfrac{2\pi}{3}\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{6}+k\pi\)
b. \(sin\left(x+\dfrac{\pi}{6}\right)\ne0\Rightarrow x+\dfrac{\pi}{6}\ne k\pi\Rightarrow x\ne-\dfrac{\pi}{6}+k\pi\)
c. \(\dfrac{1+x}{2-x}\ge0\Rightarrow-1\le x< 2\)
a) \(y=\dfrac{4}{sin^22x-1}\)
Xác định khi và chỉ khi
\(sin^22x-1\ne0\)
\(\Leftrightarrow sin^22x\ne1\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x\ne1\\sin2x\ne-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x\ne sin\dfrac{\pi}{2}\\sin2x\ne sin\dfrac{3\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x\ne\dfrac{\pi}{2}+k2\pi\\2x\ne\dfrac{3\pi}{2}+k2\pi\\2x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{3\pi}{4}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\Leftrightarrow x\ne\pm\dfrac{\pi}{4}+k\pi\)
Vậy tập xác định là \(D=R\)\\(\left\{\pm\dfrac{\pi}{4}+k\pi\right\}\)
2:
a: \(y=4+\left(cos^2x-sin^2x\right)+\left(cos^2x+sin^2x\right)\)
\(=4+1+cos2x=cos2x+5\)
-1<=cos2x<=1
=>-1+5<=cos2x+5<=1+5
=>4<=cos2x+5<=6
TGT là T=[4;6]
b: \(y=5-\dfrac{3}{2}\cdot2sinx\cdot cosx=-\dfrac{3}{2}sin2x+5\)
-1<=sin 2x<=1
=>-3/2<=-3/2sin2x<=3/2
=>-3/2+5<=y<=3/2+5
=>7/2<=y<=13/2
=>TGT là T=[7/2;13/2]
c: -1<=sin x<=1
=>-2<=-2sin x<=2
=>3<=-2sinx+5<=7
=>\(\dfrac{4}{3}>=\dfrac{4}{-2sinx+5}>=\dfrac{4}{7}\)
TGT là T=[4/7;4/3]
Đáp án D
Theo bất đẳng thức Cauchy – Schwarz có
Do đó
