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6.
Hàm số xác định khi \(\left\{{}\begin{matrix}2\sqrt{2}sinx-2\ne0\\sin3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinx\ne\dfrac{1}{\sqrt{2}}\\sin3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k2\pi\\x\ne\dfrac{3\pi}{4}+k2\pi\\x\ne\dfrac{k\pi}{3}\end{matrix}\right.\).
10.
Hàm số xác định khi \(\left\{{}\begin{matrix}sin\left(3x+\dfrac{\pi}{6}\right)\ne0\\cos2x\ne0\\sinx+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sin\left(3x+\dfrac{\pi}{6}\right)\ne0\\cos2x\ne0\\sinx+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\).
5.
\(AA'\perp\left(A'B'C'D'\right)\) theo t/c lập phương
\(\Rightarrow AA'\perp B'C'\Rightarrow\) góc giữa 2 đường thẳng bằng 90 độ
6.
\(y'=\left(x.cosx\right)'=x'.cosx+\left(cosx\right)'.x=cosx-x.sinx\)
7.
\(y'=-3x^2-5\)
\(y''=-6x\)
8.
\(\lim\limits_{x\rightarrow+\infty}\left(x^3+3x-2\right)=\lim\limits_{x\rightarrow+\infty}x^3\left(1+\dfrac{3}{x}-\dfrac{2}{x^3}\right)=+\infty.1=+\infty\)
1.
\(cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{\pi}{3}+x\right)+1=0\)
\(\Leftrightarrow2cos^2\left(\dfrac{\pi}{3}+x\right)+cos\left(\dfrac{\pi}{3}+x\right)=0\)
\(\Leftrightarrow cos\left(\dfrac{\pi}{3}+x\right)\left[2cos\left(\dfrac{\pi}{3}+x\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{3}+x\right)=0\\cos\left(\dfrac{\pi}{3}+x\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}+x=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{3}+x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)
\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)
1.
\(D=R\backslash\left\{\dfrac{\pi}{6}+\dfrac{k\pi}{3}\right\}\) là miền đối xứng
\(f\left(-x\right)=\left(-x^3-x\right)tan\left(-3x\right)=\left(x^3+x\right)tan3x=f\left(x\right)\)
Hàm chẵn
2.
\(D=R\)
\(f\left(-x\right)=\left(-2x+1\right)sin\left(-5x\right)=\left(2x-1\right)sin5x\ne\pm f\left(x\right)\)
Hàm không chẵn không lẻ
3.
\(D=R\backslash\left\{\dfrac{\pi}{6}+\dfrac{k\pi}{3}\right\}\) là miền đối xứng
\(f\left(-x\right)=tan\left(-3x\right).sin\left(-5x\right)=-tan3x.\left(-sin5x\right)=tan3x.sin5x=f\left(x\right)\)
Hàm chẵn
4.
\(D=R\)
\(f\left(-x\right)=sin^2\left(-2x\right)+cos\left(-10x\right)=sin^22x+cos10x=f\left(x\right)\)
Hàm chẵn
5.
\(D=R\backslash\left\{k\pi\right\}\) là miền đối xứng
\(f\left(-x\right)=\dfrac{-x}{sin\left(-x\right)}=\dfrac{-x}{-sinx}=\dfrac{x}{sinx}=f\left(x\right)\)
Hàm chẵn
\(\lim\limits_{x\rightarrow5}\left(x^3+5x^2-10x+8\right)=5^3+5.5^2-10.5+8=...\)
\(\lim\limits_{x\rightarrow-2}\dfrac{x^3-x^2-2x-8}{x^2+3x+2}=\dfrac{-16}{0}=-\infty\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-5x+2}{2\left|x\right|+1}=\lim\dfrac{\left|x\right|-5+\dfrac{2}{\left|x\right|}}{2+\dfrac{1}{\left|x\right|}}=\dfrac{+\infty}{2}=+\infty\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt[3]{x^3+4x-3}-4x}{\sqrt{9x^2-5x+1}-4x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(\sqrt[3]{1+\dfrac{4}{x^2}-\dfrac{3}{x^3}}-4\right)}{x\left(\sqrt[]{9-\dfrac{5}{x}+\dfrac{1}{x^2}}-4\right)}=\dfrac{1-4}{3-4}=3\)
Lời giải:
a.
\(\lim\limits_{x\to 5}(x^3+5x^2-10x+8)=5^3+5.5^2-10.5+8=208\)
b.
\(L=\lim\limits_{x\to -2}\frac{x^3-x^2-2x-8}{x^2+3x+2}\lim\limits_{x\to -2}\frac{x^3-x^2-2x-8}{x+1}.\frac{1}{x+2}=16\lim\limits_{x\to -2}\frac{1}{x+2}\)\(\lim\limits_{x\to -2-}\frac{1}{x+2}=-\infty \Rightarrow L=-\infty ; \lim\limits_{x\to -2+}\frac{1}{x+2}=+\infty \Rightarrow L=+\infty \)
b.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)
\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(x+\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\pi+k2\pi\\2x+\dfrac{\pi}{6}=-x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
c.
\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)
\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
2.
\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)
\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)
\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)
Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)
\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)
3.
\(y=\dfrac{1-sin^24x}{5}=\dfrac{cos^24x}{5}\)
\(cos4x\in\left[-1;1\right]\Rightarrow cos^24x\in\left[0;1\right]\Rightarrow y\in\left[0;\dfrac{1}{5}\right]\Rightarrow\left\{{}\begin{matrix}y_{min}=0\\y_{max}=\dfrac{1}{5}\end{matrix}\right.\)
6.
\(y=sinx+cosx+2=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\)
\(sin\left(x+\dfrac{\pi}{4}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\in\left[-\sqrt{2}+2;\sqrt{2}+2\right]\)
\(\Rightarrow y_{min}=-\sqrt{2}+2\)
\(y_{max}=\sqrt{2}+2\)