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`sin(2x-π/3)+1=0`
`<=>sin(2x-π/3)=-1`
`<=>2x-π/3=-π/2=k2π`
`<=>x=(5π)/12+kπ (k \in ZZ)`
Có: `-2020π < (5π)/12+kπ < 2020π`
`<=> -2020 < 5/12+k<2020`
`<=>-2020-5/12 <k<2020+5/12`
`=> k \in {-2020;.....;2020}`
`=>` Có `4041` giá trị của `k` thỏa mãn.
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12.
\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt[]{2}\)
\(\Rightarrow M=\sqrt{2}\)
13.
Pt có nghiệm khi:
\(5^2+m^2\ge\left(m+1\right)^2\)
\(\Leftrightarrow2m\le24\)
\(\Rightarrow m\le12\)
14.
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=k2\pi\)
15.
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(3\right)+k\pi\end{matrix}\right.\)
Đáp án A
16.
\(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
\(\left[{}\begin{matrix}2\pi\le\dfrac{\pi}{3}+k2\pi\le2018\pi\\2\pi\le\pi+k2\pi\le2018\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le k\le1008\\1\le k\le1008\end{matrix}\right.\)
Có \(1008+1008=2016\) nghiệm
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{4x-3}-\sqrt[3]{6x-5}}{x^3-x^2-x+1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[]{4x-3}-\left(2x-1\right)\right)+\left(\left(2x-1\right)-\sqrt[3]{6x-5}\right)}{x^2\left(x-1\right)-\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{4x-3-\left(2x-1\right)^2}{\sqrt[]{4x-3}+2x-1}+\dfrac{\left(2x-1\right)^3-\left(6x-5\right)}{\left(2x-1\right)^2+\left(2x-1\right)\sqrt[3]{6x-5}+\sqrt[3]{6x-5}}}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{-4\left(x-1\right)^2}{\sqrt[]{4x-3}+2x-1}+\dfrac{4\left(x-1\right)^2\left(2x+1\right)}{\left(2x-1\right)^2+\left(2x-1\right)\sqrt[3]{6x-5}+\sqrt[3]{\left(6x-5\right)^2}}}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{4}{\sqrt[]{4x-3}+2x-1}+\dfrac{4\left(2x+1\right)}{\left(2x-1\right)^2+\left(2x-1\right)\sqrt[3]{6x-5}+\sqrt[3]{\left(6x-5\right)^2}}}{x+1}=1\)