ĐKXĐ: \(x^2-4\ne0\)

=>\(x^2\ne4\)

=>\(x\notin\left\{2;-2\right\}\)

vậy: TXĐ là D=R\{2;-2}

Chọn B

1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)

 \(y=2-\left(-cosx\right).\left(-sinx\right)\)

y = 2 - sinx.cosx

y = \(2-\dfrac{1}{2}sin2x\)

Max = 2 + \(\dfrac{1}{2}\) = 2,5

Min = \(2-\dfrac{1}{2}\) = 1,5

2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)

Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)

Max = \(\sqrt{5}\)

Ta có: 

\(-1\le\sin2x\le1\)

=> \(\sqrt{4-2.\left(1\right)^5}-8\le\sqrt{4-2.\left(\sin2x\right)^5}-8\le\sqrt{4-2.\left(-1\right)^5}-8\)

=> \(\sqrt{2}-8\le\sqrt{4-2.\left(\sin2x\right)^5}-8\le\sqrt{6}-8\)

=> tìm ddc min và max

b: \(f\left(-x\right)=\dfrac{\left|-x+1\right|+\left|-x-1\right|}{\left|-x+1\right|-\left|-x-1\right|}\)

\(=\dfrac{\left|x-1\right|+\left|x+1\right|}{\left|x-1\right|-\left|x+1\right|}\)

=-f(x)

Vậy: f(x) là hàm số lẻ

\(y'=\dfrac{1}{5}\cdot5x^4+\dfrac{1}{2\sqrt{x}}+\dfrac{1}{x^2}+\dfrac{-\left(x^3\right)'}{x^6}\)

\(=x^4+\dfrac{1}{2\sqrt{x}}+\dfrac{1}{x^2}+\dfrac{-3x^2}{x^6}=x^4+\dfrac{1}{2\sqrt{x}}+\dfrac{1}{x^2}-\dfrac{3}{x^4}\)

Em kiểm tra đề là \(\dfrac{y^2}{4}\) hay \(\dfrac{y^4}{4}\)

Nếu đề đúng là \(\dfrac{y^4}{4}\) thì có thể coi như là không giải được

\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2-xy+\dfrac{y^2}{4}\right)+xy=2\)

\(\Leftrightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2+xy\ge xy\)

\(\Rightarrow P_{max}=2023\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;-2\right);\left(1;2\right)\)

\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2+xy+\dfrac{y^2}{4}\right)-xy=2\)

\(\Rightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x+\dfrac{y}{2}\right)^2-xy\ge-xy\)

\(\Rightarrow xy\ge-2\Rightarrow P\ge2019\)

\(P_{min}=2019\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x+\dfrac{y}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;2\right);\left(1;-2\right)\)

2.Biểu thức luôn xác định

\(y=\dfrac{4}{\sqrt{5-2cos^2sin^2x}}=\dfrac{4}{\sqrt{5-\dfrac{1}{2}sin^22x}}\)

Có: \(1\ge sin^22x\ge0\)

\(\Leftrightarrow-\dfrac{1}{2}\le-\dfrac{1}{2}sin^22x\le0\)

\(\Leftrightarrow\dfrac{3\sqrt{2}}{2}\le\sqrt{5-\dfrac{1}{2}sin^22x}\le\sqrt{5}\)

\(\Rightarrow\dfrac{4\sqrt{2}}{3}\ge y\ge\dfrac{4\sqrt{5}}{5}\)

miny=\(\dfrac{4\sqrt{5}}{5}\) \(\Leftrightarrow sin2x=0\)\(\Leftrightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\)

maxy=\(\dfrac{4\sqrt{2}}{3}\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{-\pi}{4}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

1.Biểu thức luôn xác định

Xét \(sin2x=0\) \(\Leftrightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\) khi đó \(y=-6\)

Xét \(sin2x\ne0\) 

=> \(1\ge sin^52x\ge-1\)

\(\Leftrightarrow4-1\le4-sin^52x\le4+1\)

\(\Leftrightarrow\sqrt{3}\le\sqrt{4-sin^52x}\le\sqrt{5}\)

\(\Leftrightarrow\sqrt{3}-8\le y\le\sqrt{5}-8\)

\(y=\sqrt{3}-8< -6\) , \(y=\sqrt{5}-8>-6\)

=>min= \(\sqrt{3}-8\) \(\Leftrightarrow sin2x=1\left(tm\right)\) \(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

maxy=\(\sqrt{5}-8\)\(\Leftrightarrow sin2x=-1\left(tm\right)\) \(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

(câu này e ko chắc)

Chọn D

Lời giải:
\(x\in [-\sqrt{2}; \sqrt{2}]\Rightarrow x^2\leq 2\Rightarrow \sqrt{x^2+1}\leq \sqrt{3}\)

\(y=\frac{x+1}{\sqrt{x^2+1}}\geq \frac{x+1}{\sqrt{3}}\geq \frac{-\sqrt{2}+1}{\sqrt{3}}\)

Vậy $y_{\min}=\frac{-\sqrt{2}+1}{\sqrt{3}}$ khi $x=-\sqrt{2}$

$y^2=\frac{x^2+2x+1}{x^2+1}=1+\frac{2x}{x^2+1}$

$y^2=2+\frac{2x-x^2-1}{x^2+1}=2-\frac{(x-1)^2}{x^2+1}\leq 2$

$\Rightarrow y\leq \sqrt{2}$

Vậy $y_{\max}=\sqrt{2}$ khi $x=1$