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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:
\(\left(m-2\right)^2+\left(2m+1\right)^2< 9m^2\)
\(\Leftrightarrow m^2-4m+4+4m^2+4m+1< 9m^2\)
\(\Leftrightarrow m^2-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
Đ á p á n B P T đ ã c h o t ư ơ n g đ ư ơ n g : 4 . cos 2 2 x + 8 sin 2 x - 7 = 0 ⇔ 4 . 1 - sin 2 2 x + 8 . sin 2 x - 7 = 0 ⇔ - 4 . sin 2 2 x + 8 . sin 2 x - 3 = 0 ⇔ sin 2 x = 1 2 ⇔ x = π 12 + k π ( k ∈ ℤ ) hoặc x = 5 π 12 + kπ ( k ∈ ℤ )
