1: Ta có: \(-1<=\sin\left(2x+\frac{\pi}{4}\right)\le1\)

=>\(-3\le3\cdot\sin\left(2x+\frac{\pi}{4}\right)\le3\)

=>\(-3-1\le3\cdot\sin\left(2x+\frac{\pi}{4}\right)-1\le3-1\)

=>-4<=y<=2

=>Tập giá trị là T=[-4;2]

\(y_{\min}=-4\) khi \(\sin\left(2x+\frac{\pi}{4}\right)=-1\)

=>\(2x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\)

=>\(2x=-\frac34\pi+k2\pi\)

=>\(x=-\frac38\pi+k\pi\)

2: \(0\le cos^2x\le1\)

=>\(0\ge-5\cdot cos^2x\ge-5\)

=>\(0+3\ge-5\cdot cos^2x+3\ge-5+3\)

=>3>=y>=-2

=>Tập giá trị là T=[-2;3]

\(y_{\max}=3\) khi \(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

\(y_{\min}=-2\) khi \(cos^2x=0\)

=>cosx=0

=>\(x=\frac{k\pi}{2}\)

3: \(-1\le cosx\le1\)

=>\(-3\le3\cdot cosx\le3\)

=>\(-3+4\le3\cdot cosx+4\le3+4\)

=>\(1\le3\cdot cosx+4\le7\)

=>\(\frac51\ge\frac{5}{3\cdot cosx+4}\ge\frac57\)

=>\(\frac57\le y\le5\)

=>Tập giá trị là \(T=\left\lbrack\frac57;5\right\rbrack\)

\(y_{\min}=\frac57\) khi cosx=1

=>\(x=k2\pi\)

\(y_{\max}=5\) khi cosx=-1

=>\(x=\pi+k2\pi\)

4: \(y=\sin^2x-4\cdot\sin x+8\)

\(=\sin^2x-4\cdot\sin x+4+4\)

\(=\left(\sin x-2\right)^2+4\)

Ta có: \(-1\le\sin x\le1\)

=>\(-1-2\le\sin x-2\le1-2\)

=>\(-3\le\sin x-2\le-1\)

=>\(1\le\left(\sin x-2\right)^2\le9\)

=>\(5\le\left(\sin x-2\right)^2+4\le13\)

=>5<=y<=13

=>Tập giá trị là T=[5;13]

\(y_{\min}=5\) khi sin x-2=-1

=>sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

\(y_{\max}\) =13 khi sin x-2=-3

=>sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

\(a,D=R\\ b,2x-3>0\\ \Rightarrow x>\dfrac{3}{2}\\ \Rightarrow D=(\dfrac{3}{2};+\infty)\\ c,-x^2+4>0\\ \Rightarrow x^2< 4\\ \Leftrightarrow-2< x< 2\\ \Rightarrow D=\left(-2;2\right)\)

1.

\(y'=12x+\dfrac{4}{x^2}\)

2.

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

3.

\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)

4.

\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)

\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)

5.

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

6.

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

a) y' = 5x⁴ - 12x² + 2.

b) y' = - + 2x - 2x³.

c) y' = 2x³ - 2x² + .

d) y = 24x⁵ - 9x⁷ => y' = 120x⁴ - 63x⁶.

1: Để đây là cấp số cộng thì

\(\left\{{}\begin{matrix}x=\dfrac{2+8}{2}=\dfrac{10}{2}=5\\8=\dfrac{y+x}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y+5=16\end{matrix}\right.\)

=>x=5 và y=11

2: Để bốn số này là cấp số cộng thì

\(\left\{{}\begin{matrix}a+b+1=2\cdot5=10\\b+1=\dfrac{5+13}{2}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=8\\a=1\end{matrix}\right.\)

1/ a/ \(y'=-5sinx+\frac{3}{cos^2\left(x+15^0\right)}\)

b/ \(y'=\frac{6cos3x\left(-4cosx-5\right)-8sinx.sin3x}{\left(4cosx+5\right)^2}\)

2/ \(y'=-3cosx-\frac{15}{sin^23x}\Rightarrow y'\left(\frac{\pi}{4}\right)=-3cos\left(\frac{\pi}{4}\right)-\frac{15}{sin^2\left(\frac{3\pi}{4}\right)}=-\frac{60+3\sqrt{2}}{2}\)

3/ \(y'=4x-5\)

a/ \(y'\left(2\right)=3\) ; \(y\left(2\right)=2\)

Tiếp tuyến: \(y=3\left(x-2\right)+2=3x-4\)

b/ Tiếp tuyến song song \(y=2x-3\Rightarrow\) có hệ số góc bằng 2

\(\Rightarrow4x_0-5=2\Rightarrow x_0=\frac{7}{4}\Rightarrow y\left(\frac{7}{4}\right)=\frac{11}{8}\)

Tiếp tuyến: \(y=2\left(x-\frac{7}{4}\right)+\frac{11}{8}\)

c/ \(-x+3y-1=0\Rightarrow y=\frac{1}{3}x+\frac{1}{3}\)

Tiếp tuyến vuông góc với d nên có hệ số góc bằng \(-3\)

\(\Rightarrow4x_0-5=-3\Rightarrow x_0=\frac{1}{2}\Rightarrow y\left(\frac{1}{2}\right)=2\)

Tiếp tuyến: \(y=-3\left(x-\frac{1}{2}\right)+2\)

3/ a, y=\(2x^2-5x+4\)

Ta có: \(x_o=2\)-> \(y_0=2\)

-> \(f'\left(x_0\right)=3\)

Nên ta có pttt: y'= 3x - 4

a.

\(y=\dfrac{3}{2}sin2x-2\left(cos^2x-sin^2x\right)+5=\dfrac{3}{2}sin2x-2cos2x+5\)

\(=\dfrac{5}{2}\left(\dfrac{3}{5}sin2x-\dfrac{4}{5}cos2x\right)+5=\dfrac{5}{2}sin\left(2x-a\right)+5\) (với \(cosa=\dfrac{3}{5}\))

\(\Rightarrow-\dfrac{5}{2}+5\le y\le\dfrac{5}{2}+5\)

b.

\(\Leftrightarrow y.sinx-2y.cosx+4y=3sinx-cosx+1\)

\(\Leftrightarrow\left(y-3\right)sinx+\left(1-2y\right)cosx=1-4y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(y-3\right)^2+\left(1-2y\right)^2\ge\left(1-4y\right)^2\)

\(\Leftrightarrow11y^2+2y-9\le0\)

\(\Leftrightarrow-1\le y\le\dfrac{9}{11}\)

c.

Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)

\(\Rightarrow y=\dfrac{2\left(sin^2a+6sina.cosa\right)}{1+2sina.cosa+cos^2a}=\dfrac{1-cos2a+6sin2a}{1+sin2a+\dfrac{1+cos2a}{2}}=\dfrac{2-2cos2a+12sin2a}{3+2sin2a+cos2a}\)

\(\Leftrightarrow3y+2y.sin2a+y.cos2a=2-2cos2a+12sin2a\)

\(\Leftrightarrow\left(2y-12\right)sin2a+\left(y+2\right)cos2a=2-3y\)

Theo điều kiện có nghiệm của pt bậc nhất theo sin2a, cos2a:

\(\left(2y-12\right)^2+\left(y+2\right)^2\ge\left(2-3y\right)^2\)

\(\Leftrightarrow y^2+8y-36\le0\)

\(\Rightarrow-4-2\sqrt{13}\le y\le-4+2\sqrt{13}\)