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

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

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

\(y=\sin^4x+\cos^4x\\ =1-2\sin^2x\cdot\cos^2x\\ =1-\dfrac{1}{2}\sin^22x\\ 0\le\sin^22x\le1\\ \Leftrightarrow\dfrac{1}{2}\le y\le1\\ y_{min}=\dfrac{1}{2}\Leftrightarrow\sin^22x=1\Leftrightarrow x=\dfrac{k\pi}{2}\pm\dfrac{\pi}{4}\\ y_{max}=1\Leftrightarrow\sin^22x=0\Leftrightarrow x=k\pi\)

\(y=3\sin x+4\cos x\\ =5\left(\dfrac{3\sin x}{5}+\dfrac{4\cos x}{5}\right)\\ =5\cos\left(x-a\right),\forall\cos a=\dfrac{4}{5},\sin a=\dfrac{3}{5}\\ -1\le\cos\left(x-a\right)\le1\\ \Leftrightarrow-5\le y\le5\\ y_{min}=-5\Leftrightarrow\cos\left(x-a\right)=-1\\ y_{max}=5\Leftrightarrow\cos\left(x-a\right)=1\)

Em không thấy đáp án giống như trên lời giải, có thể giúp em làm cách khác không ạ?

Bạn Phúc hơi nhầm 1 xíu

\(y=4sinx\left(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)=2sinx.cosx-2\sqrt{3}sin^2x\)

\(=sin2x-\sqrt{3}\left(1-cos2x\right)=sin2x+\sqrt{3}cos2x-\sqrt{3}\)

\(=2\left(\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\right)-\sqrt{3}\)

\(=2cos\left(2x-\dfrac{\pi}{6}\right)-\sqrt{3}\)

\(\Rightarrow y_{min}=-2-\sqrt{3}\) ; \(y_{max}=2-\sqrt{3}\)

Đáp án mà đề đưa ra như bên dưới đều sai cả.

\(y'=1-\sqrt{2}\sin x=\dfrac{1}{\sqrt{2}}\Rightarrow x=\dfrac{\pi}{4}\\ y\left(0\right)=\sqrt{2};y\left(\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+1;y\left(\dfrac{\pi}{2}\right)=\dfrac{\pi}{2}\\ \Rightarrow y_{max}=y\left(\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+1\\ y_{min}=y\left(0\right)=\sqrt{2}\)

Đáp án đúng : D

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

Đáp án đúng : C

2.

$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

a.

Đặt \(cos2x=t\Rightarrow t\in\left[-1;1\right]\)

Xét hàm \(y=f\left(t\right)=2t^2+2t-4\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=-\dfrac{1}{2}\in\left[-1;1\right]\)

\(f\left(-1\right)=-4\) ; \(f\left(-\dfrac{1}{2}\right)=-\dfrac{9}{2}\) ; \(f\left(1\right)=0\)

\(\Rightarrow y_{min}=-\dfrac{9}{2}\) khi \(t=-\dfrac{1}{2}\) hay \(cos2x=-\dfrac{1}{2}\)

\(y_{max}=0\) khi \(cos2x=1\)

b. Đặt \(tanx=t\Rightarrow t\in\left[-1;\sqrt{3}\right]\)

Xét hàm \(f\left(t\right)=t^2-2\sqrt{3}t-1\) trên \(\left[-1;\sqrt{3}\right]\)

\(-\dfrac{b}{2a}=\sqrt{3}\in\left[-1;\sqrt{3}\right]\)

\(f\left(-1\right)=2\sqrt{3}\) ; \(f\left(\sqrt{3}\right)=-4\)

\(y_{min}=-4\) khi \(x=\dfrac{\pi}{3}\) ; \(y_{max}=2\sqrt{3}\) khi \(x=-\dfrac{\pi}{4}\)