Question

[Anni Nguyễn]

giải hộ mình câu 5,6,7 với ạ

Answer 1

[Lê Thị Thục Hiền]

Câu 5:

\(y=1-\left(sin2x+cos2x\right)^3\)

\(=1-\left[\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right]^3\)\(=1-2\sqrt{2}.sin^3\left(2x+\dfrac{\pi}{4}\right)\)

Có \(-1\le sin\left(2x+\dfrac{\pi}{4}\right)\le1\)

\(\Leftrightarrow-1\le sin^3\left(2x+\dfrac{\pi}{4}\right)\le1\) \(\Leftrightarrow1+2\sqrt{2}\ge y\ge1-2\sqrt{2}\)

\(\Rightarrow y_{min}=1-2\sqrt{2}\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow x=\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)

\(\Rightarrow y_{max}=1+2\sqrt{2}\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=-1\)\(\Leftrightarrow x=\dfrac{-3\pi}{8}+k\pi\left(k\in Z\right)\)

Ý B

Answer 2

[Lê Thị Thục Hiền]

Câu 6: Hàm số có TXĐ: D=R

\(y=\sqrt{4-2sin^52x}-8\) 

Có \(-1\le sin2x\le1\)

\(\Leftrightarrow-1\le sin^52x\le1\)

\(\Leftrightarrow2\ge-2sin^52x\ge-2\)

\(\Leftrightarrow\)\(\sqrt{6}-8\ge y\ge\sqrt{2}-8\) 

Ý A

Câu 7: TXĐ: D=R

\(y=\dfrac{3}{3-\sqrt{1-cosx}}\)

Có \(-1\le cosx\le1\) \(\Leftrightarrow2\ge1-cosx\ge0\) \(\Leftrightarrow3-\sqrt{2}\ge3-\sqrt{1-cosx}\ge3\)

\(\Leftrightarrow\dfrac{3}{3-\sqrt{2}}\le y\le1\)

Vậy \(y_{min}=\dfrac{3}{3-\sqrt{2}}\Leftrightarrow x=\pi+k2\pi\) (k nguyên)

\(y_{max}=1\Leftrightarrow x=k2\pi\)  (k nguyên)