\(y=\sqrt{5}\left(\dfrac{1}{\sqrt{5}}cosx-\dfrac{2}{\sqrt{5}}sinx\right)+3\)
Đặt \(\dfrac{1}{\sqrt{5}}=cos\alpha\Rightarrow\dfrac{2}{\sqrt{5}}=sin\alpha\)
\(\Rightarrow y=\sqrt{5}\left(cosx.cos\alpha-sinx.sin\alpha\right)+3=\sqrt{5}cos\left(x+\alpha\right)+3\)
Do \(-1\le cos\left(x+\alpha\right)\le1\)
\(\Rightarrow3-\sqrt{5}\le y\le3+\sqrt{5}\)
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