24.

\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

\(y_{max}=4\)

26.

\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)

Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)

\(y_{max}=\sqrt{2}\)

b.

\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Đáp án đúng : D

Ta có: \(y=\sqrt{3+x}+\sqrt{5-x}\)

ĐKXĐ: \(-3\le x\le5\)

\(y^2=3+x+5-x+2\sqrt{\left(3+x\right)\left(5-x\right)}=8+2\sqrt{\left(3+x\right)\left(5-x\right)}\)\(\ge8\)

\(\Rightarrow y\ge2\sqrt{2}\)

Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)(thỏa mãn)

Vậy min y = \(2\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

mặt khác \(y^2\) = \(8+2\sqrt{\left(3+x\right)\left(5-x\right)}\le8+3+x+5-x=16\)

\(\Rightarrow y\le4\)

Dấu"=" xảy ra khi và chỉ khi \(3+x=5-x\Leftrightarrow x=1\)(thỏa mãn)

Vậy max y = 4 \(\Leftrightarrow x=1\)

\(y=\left(x+2\right)\left(3-x\right)\)

\(=3x-x^2+6-2x\)

\(=-x^2+x+6\)

=>y'=-2x+1

Đặt y'=0

=>-2x+1=0

=>-2x=-1

=>\(x=\dfrac{1}{2}\)

\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+2\right)\left(3-\dfrac{1}{2}\right)=\dfrac{5}{2}\cdot\dfrac{5}{2}=\dfrac{25}{4}\)

\(f\left(-2\right)=\left(-2+2\right)\left(3+2\right)=0\)

\(f\left(3\right)=\left(3+2\right)\left(3-3\right)=0\)

=>\(y_{max\left[-2;3\right]}=\dfrac{25}{4}\)

`TXĐ: R`

Ta có: `-1 <= sin(x+ \pi/3) <= 1`

`<=>0 <= sin^4 (x+\pi/3) <= 1`

`<=>2 <= y <= 3`

    `=>y_[mi n]=2<=>sin(x +\pi/3)=0<=>x= -\pi/3+k\pi`   `(k in ZZ)`

        `y_[max]=3<=>sin(x +\pi/3)=1<=>x=\pi/6 +k2\pi`  `(k in ZZ)`