Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

\(lim\left(\sqrt{x^2-x+1}\right)-x\)

\(limx\left[\left(\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}\right)-1\right]\)

\(=x\left(\sqrt{1}-1\right)=0\)

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-x+1}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(\sqrt{x^2-x+1}-x\right)\left(\sqrt{x^2-x+1}+x\right)}{\sqrt{x^2-x+1}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-x+1}{\sqrt{x^2-x+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{-1+\dfrac{1}{x}}{\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}+1}=-\dfrac{1}{2}\)

\(32cos^6\left(\dfrac{x}{2}\right)+3sinx-4sin^3x=3sinx\)

\(\Leftrightarrow\left(2cos^2\dfrac{x}{2}\right)^3=sin^3x\)

\(\Leftrightarrow2cos^2\dfrac{x}{2}=sinx\)

\(\Leftrightarrow cosx+1=sinx\)

\(\Leftrightarrow sinx-cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow...\)

\(\Leftrightarrow4sin^{2020}x\left(1-2sin^2x\right)=4cos^{2020}x\left(2cos^2x-1\right)+5cos2x=0\)

\(\Leftrightarrow4sin^{2020}x.cos2x=4cos^{2020}x.cos2x+5cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=...\\4sin^{2020}x=4cos^{2020}x+5\left(1\right)\end{matrix}\right.\)

Xét (1), ta có \(\left\{{}\begin{matrix}4sin^{2020}x\le4\\4cos^{2020}x+5\ge5\end{matrix}\right.\)

\(\Rightarrow4sin^{2020}x< 4cos^{2020}x+5\) với mọi x

\(\Rightarrow\left(1\right)\) vô nghiệm

\(32sin^6\dfrac{x}{2}+sin3x=3sinx\)

\(\Leftrightarrow32sin^6\dfrac{x}{2}+3sinx-4sin^3x=3sinx\)

\(\Leftrightarrow8sin^6\dfrac{x}{2}=sin^3x\)

\(\Leftrightarrow8sin^6\dfrac{x}{2}=8sin^3\dfrac{x}{2}.cos^3\dfrac{x}{2}\)

\(\Leftrightarrow sin^3\dfrac{x}{2}\left(1-cos^3\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{x}{2}=0\\cos\dfrac{x}{2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=k\pi\\\dfrac{x}{2}=k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=k4\pi\end{matrix}\right.\)

\(\Leftrightarrow x=k2\pi\)

Bn tk nha:

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{2}\right)=\dfrac{1}{2}+\dfrac{1}{2}cos2x\)

\(\Leftrightarrow-sin2x=cos2x\)

\(\Rightarrow tan2x=-1\)

\(\Rightarrow2x=-\dfrac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\)

\(A=\dfrac{\sqrt{x^3+y^3+1}}{xy}+\dfrac{\sqrt{y^3+z^3+1}}{yz}+\dfrac{\sqrt{z^3+x^3+1}}{zx}\)

\(\dfrac{\sqrt{x^3+y^3+1}}{xy}=\dfrac{\sqrt{x^3+y^3+xyz}}{xy}\ge\dfrac{\sqrt{xy\left(x+y\right)+xyz}}{xy}=\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}\ge\dfrac{\sqrt{xy.3^3\sqrt{xyz}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)

\(\dfrac{\sqrt{y^3+z^3+1}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}}\)

\(\dfrac{\sqrt{z^3+x^3+1}}{zx}\ge\dfrac{\sqrt{3}}{\sqrt{zx}}\)

\(\Rightarrow A\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.xz}}}=3\sqrt{3}.\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

\(\Leftrightarrow3\left|sinx\right|=2-cosx\)

\(\Leftrightarrow9sin^2x=4-4cosx+cos^2x\)

\(\Leftrightarrow9-9cos^2x=4-4cosx+cos^2x\)

\(\Leftrightarrow10cos^2x-4cosx-5=0\)

\(\Leftrightarrow...\)

Câu 11: B

Câu 12: A

Câu 13: C

Câu 14: B

A