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

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

=>\(y'=1+\dfrac{4'\left(x-2\right)^2-4\left[\left(x-2\right)^2\right]'}{\left(x-2\right)^4}\)

=>\(y'=1+\dfrac{-4\cdot2\cdot\left(x-2\right)'\left(x-2\right)}{\left(x-2\right)^4}\)

=>\(y'=1-\dfrac{8}{\left(x-2\right)^3}\)

Đặt y'=0

=>\(\dfrac{8}{\left(x-2\right)^3}=1\)

=>\(\left(x-2\right)^3=8\)

=>x-2=2

=>x=4

Đặt \(f\left(x\right)=x+\dfrac{4}{\left(x-2\right)^2}\)

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

\(f\left(0\right)=0+\dfrac{4}{\left(0-2\right)^2}=0+\dfrac{4}{4}=1\)

\(f\left(5\right)=5+\dfrac{4}{\left(5-2\right)^2}=5+\dfrac{4}{9}=\dfrac{49}{9}\)

Vì f(0)<f(4)<f(5)

nên \(f\left(x\right)_{max\left[0;5\right]\backslash\left\{2\right\}}=f\left(5\right)=\dfrac{49}{9}\) và \(f\left(x\right)_{min\left[0;5\right]\backslash\left\{2\right\}}=1\)

2: \(y=cos^22x-sinx\cdot cosx+4\)

\(=1-sin^22x-\dfrac{1}{2}\cdot sin2x+4\)

\(=-sin^22x-\dfrac{1}{2}\cdot sin2x+5\)

\(=-\left(sin^22x+\dfrac{1}{2}\cdot sin2x-5\right)\)

\(=-\left(sin^22x+2\cdot sin2x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{81}{16}\right)\)

\(=-\left(sin2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}\)

\(-1< =sin2x< =1\)

=>\(-\dfrac{3}{4}< =sin2x+\dfrac{1}{4}< =\dfrac{5}{4}\)

=>\(0< =\left(sin2x+\dfrac{1}{4}\right)^2< =\dfrac{25}{16}\)

=>\(0>=-\left(sin2x+\dfrac{1}{4}\right)^2>=-\dfrac{25}{16}\)

=>\(\dfrac{81}{16}>=-sin\left(2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}>=-\dfrac{25}{16}+\dfrac{81}{16}=\dfrac{7}{2}\)

=>\(\dfrac{81}{16}>=y>=\dfrac{7}{2}\) 

\(y_{min}=\dfrac{7}{2}\) khi \(sin2x+\dfrac{1}{4}=\dfrac{5}{4}\)

=>\(sin2x=1\)

=>\(2x=\dfrac{\Omega}{2}+k2\Omega\)

=>\(x=\dfrac{\Omega}{4}+k\Omega\)

\(y_{max}=\dfrac{81}{16}\) khi sin 2x=-1

=>\(2x=-\dfrac{\Omega}{2}+k2\Omega\)

=>\(x=-\dfrac{\Omega}{4}+k\Omega\)

1.

\(y=\frac{1}{2}sin2x-1\)

Do \(-1\le sin2x\le1\Rightarrow-\frac{3}{2}\le y\le-\frac{1}{2}\)

\(y_{min}=-\frac{3}{2}\) ; \(y_{max}=-\frac{1}{2}\)

2.

\(y=5+5\left(\frac{4}{5}cosx-\frac{3}{5}sinx\right)=5+5cos\left(x+a\right)\) với \(cosa=\frac{4}{5}\)

Do \(-1\le cos\left(x+a\right)\le1\Rightarrow0\le y\le10\)

\(y_{min}=0\) ; \(y_{max}=10\)

ĐKXĐ: 

\(sinx.cosx+2sinx-cosx-2\ge0\)

\(\Leftrightarrow sinx\left(cosx+2\right)-\left(cosx+2\right)\ge0\)

\(\Leftrightarrow\left(sinx-1\right)\left(cosx+2\right)\ge0\)

\(\Leftrightarrow sinx-1\ge0\) (do \(cosx+2>0\) với mọi x)

\(\Rightarrow sinx=1\)

\(\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

Bạn chú ý viết đề bài bằng công thức toán.

Phần a là \(\sqrt{\frac{\sin x+3}{2}}\) hay\(\sqrt{\sin x+\frac{3}{2}}\)?

Phần a ạ

\(y=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).sinx.cosx\)

\(=\left(cos^2x-sin^2x\right).\dfrac{1}{2}\left(2sinx.cosx\right)=\dfrac{1}{2}cos2x.sin2x\)

\(=\dfrac{1}{4}sin4x\)

Do \(-1\le sin4x\le1\Rightarrow-\dfrac{1}{4}\le y\le\dfrac{1}{4}\)

\(y_{min}=-\dfrac{1}{4}\) khi \(sin4x=1\)

\(y_{max}=\dfrac{1}{4}\) khi \(sin4x=1\)

 +\(\sin^2x+\cos^2x=1\Leftrightarrow\sin^2x+4\sin^2x=1\Rightarrow\sin^2x=\frac{1}{5}\)

=>Sinx.cosx=sinx.2sinx=2sin²x =2.1/5 = 2/5