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câu 1:sin2x+cos2x=\(\dfrac{m}{2}\)
⇔sin2x +\(\dfrac{1+cos2x}{2}=\dfrac{m}{2}\)⇔2sin2x+cos2x=m-1
để phương trình có nghiệm thì:a2+b2≥c2⇔ 22+12≥(m-1)2⇔5≥m2-2m+1⇔m2-2m-4≤0 ⇔1-√5≤m≤1+√5. Chúc bạn học tốt.
1.
a, Phương trình có nghiệm khi:
\(\left(m+2\right)^2+m^2\ge4\)
\(\Leftrightarrow m^2+4m+4+m^2\ge4\)
\(\Leftrightarrow2m^2+4m\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\ge0\\m\le-2\end{matrix}\right.\)
b, Phương trình có nghiệm khi:
\(m^2+\left(m-1\right)^2\ge\left(2m+1\right)^2\)
\(\Leftrightarrow2m^2+6m\le0\)
\(\Leftrightarrow-3\le m\le0\)
2.
a, Phương trình vô nghiệm khi:
\(\left(2m-1\right)^2+\left(m-1\right)^2< \left(m-3\right)^2\)
\(\Leftrightarrow4m^2-4m+1+m^2-2m+1< m^2-6m+9\)
\(\Leftrightarrow4m^2-7< 0\)
\(\Leftrightarrow-\dfrac{\sqrt{7}}{2}< m< \dfrac{\sqrt{7}}{2}\)
b, \(2sinx+cosx=m\left(sinx-2cosx+3\right)\)
\(\Leftrightarrow\left(m-2\right)sinx-\left(2m+1\right)cosx=-3m\)
Phương trình vô nghiệm khi:
\(\left(m-2\right)^2+\left(2m+1\right)^2< 9m^2\)
\(\Leftrightarrow m^2-4m+4+4m^2+4m+1< 9m^2\)
\(\Leftrightarrow m^2-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)
1.
\(\lim\left(\sqrt{9^n-2.3^n}-3^n+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{\left(\sqrt{9^n-2.3^n}-3^n\right)\left(\sqrt{9^n-2.3^n}+3^n\right)}{\sqrt{9^n-2.3^n}+3^n}+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{-2.3^n}{\sqrt{9^n-2.3^n}+3^n}+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{-2.3^n}{3^n\left(\sqrt{1-\dfrac{2}{3^n}}+1\right)}+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{-2}{\sqrt{1-\dfrac{2}{3^n}}+1}+\dfrac{1}{2021}\right)\)
\(=\dfrac{-2}{1+1}+\dfrac{1}{2021}=-\dfrac{2020}{2021}\)
2.
\(AP=4PB=4\left(AB-AP\right)=4AB-4AP\)
\(\Rightarrow5AP=4AB\Rightarrow AP=\dfrac{4}{5}AB\)
\(\Rightarrow\overrightarrow{AP}=\dfrac{4}{5}\overrightarrow{AB}\)
\(CD=5CQ=5\left(CD-DQ\right)\Rightarrow5DQ=4CD\Rightarrow DQ=\dfrac{4}{5}CD\)
\(\Rightarrow\overrightarrow{DQ}=-\dfrac{4}{5}\overrightarrow{CD}\)
Ta có:
\(\overrightarrow{PQ}=\overrightarrow{PA}+\overrightarrow{AD}+\overrightarrow{DQ}=-\dfrac{4}{5}\overrightarrow{AB}+\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CD}\)
\(=-\dfrac{4}{5}\left(\overrightarrow{AD}+\overrightarrow{DB}\right)+\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CD}=-\dfrac{4}{5}\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{DB}+\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CD}\)
\(=\dfrac{1}{5}\overrightarrow{AD}-\dfrac{4}{5}\left(\overrightarrow{CD}+\overrightarrow{DB}\right)=\dfrac{1}{5}\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CB}\)
\(=\dfrac{1}{5}\overrightarrow{AD}+\dfrac{4}{5}\overrightarrow{BC}\)
Mà \(\overrightarrow{AD};\overrightarrow{BC}\) không cùng phương\(\Rightarrow\overrightarrow{AD};\overrightarrow{BC};\overrightarrow{PQ}\) đồng phẳng
c.
\(\Leftrightarrow sin4x=sin\left(3x-\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=3x-\dfrac{\pi}{2}+k2\pi\\4x=\dfrac{3\pi}{2}-3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+k2\pi\\x=\dfrac{3\pi}{14}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
d.
\(\Leftrightarrow sin\left(2x+30^0\right)=sin\left(30^0+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+30^0=30^0+x+k360^0\\2x+30^0=150^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k360^0\\x=40^0+k120^0\end{matrix}\right.\)
e.
\(\Leftrightarrow cos3x=-sinx\)
\(\Leftrightarrow cos3x=cos\left(\dfrac{\pi}{2}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+x+k2\pi\\3x=-\dfrac{\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
f.
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)\left(sin2x+cos5x\right)=0\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)\left(sin2x-sin\left(5x-\dfrac{\pi}{2}\right)\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x-\dfrac{\pi}{4}\right)=0\\sin\left(5x-\dfrac{\pi}{2}\right)=sin2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=k\pi\\5x-\dfrac{\pi}{2}=2x+k2\pi\\5x-\dfrac{\pi}{2}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{3\pi}{14}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
1.
\(u_{n+1}=4u_n+3.4^n\)
\(\Leftrightarrow u_{n+1}-\dfrac{3}{4}\left(n+1\right).4^{n+1}=4\left[u_n-\dfrac{3}{4}n.4^n\right]\)
Đặt \(u_n-\dfrac{3}{4}n.4^n=v_n\Rightarrow\left\{{}\begin{matrix}v_1=2-\dfrac{3}{4}.4=-1\\v_{n+1}=4v_n\end{matrix}\right.\)
\(\Rightarrow v_n=-1.4^{n-1}\)
\(\Rightarrow u_n=\dfrac{3}{4}n.4^n-4^{n-1}=\left(3n-1\right)4^{n-1}\)
2.
\(a_n=\dfrac{a_{n-1}}{2n.a_{n-1}+1}\Rightarrow\dfrac{1}{a_n}=2n+\dfrac{1}{a_{n-1}}\)
\(\Leftrightarrow\dfrac{1}{a_n}-n^2-n=\dfrac{1}{a_{n-1}}-\left(n-1\right)^2-\left(n-1\right)\)
Đặt \(\dfrac{1}{a_n}-n^2-n=b_n\Rightarrow\left\{{}\begin{matrix}b_1=2-1-1=0\\b_n=b_{n-1}=...=b_1=0\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a_n}=n^2+n\Rightarrow a_n=\dfrac{1}{n^2+n}\)
\(msinx+\sqrt{3}cosx=m+1\)
Phương trình có nghiệm\(\Leftrightarrow\) \(m^2+\left(\sqrt{3}\right)^2\ge\left(m+1\right)^2\)
\(\Rightarrow3\ge2m+1\Rightarrow m\le1\)
Vậy pt vô nghiệm\(\Leftrightarrow m>1\)