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 = \(\sqrt{5}\)

a.

\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)

b.

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

c.

\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)

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.

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

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

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

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

Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.

1.

\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)

\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)

2.

\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)

3.

\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)

\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)

4.

\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

5.

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

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)

8.

\(y=tan^33x-\left(sin2x+cos3x\right)^5\)

\(\Rightarrow y'=3tan^23x.\left(tan3x\right)'-5\left(sin2x+cos3x\right)^4.\left(sin2x+cos3x\right)'\)

\(=\dfrac{9.tan^23x}{cos^23x}-5\left(sin2x+cos3x\right)^4.\left(2cos2x-3sin3x\right)\)

9.

\(y'=6cot^55x.\left(cot5x\right)'-4cos^33x.\left(cos3x\right)'+3cos3x\)

\(=-\dfrac{30.cot^55x}{sin^25x}+12cos^33x.sin3x+3cos3x\)

1.

Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng

2.a

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)

\(\Leftrightarrow1+tan^2x=4tanx+6\)

\(\Leftrightarrow tan^2x-4tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)

2b.

Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)

\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin^3t=sint+cost\)

\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)

\(\Leftrightarrow sint.cos^2t+cost=0\)

\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

2c.

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)

\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)

\(\Leftrightarrow2sinx.cos2x+cos2x=1\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)

\(\Leftrightarrow1-2sin^2x-sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

a.

\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)

\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)

\(\Rightarrow y_{min}=y\left(1\right)=0\)

\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)

b.

\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]

\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)

c.

\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)

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

\(y=f\left(t\right)=-t^2-t+2\)

\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)

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

\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)

d.

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

\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)

\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)

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

1. Hàm số xác định `<=> 1-cosx \ne 0<=>cosx \ne 1<=>x \ne k2π`

Vì: `1+cosx >=0 forallx ; 1-cosx >=0 forall x`

2. Hàm số xác định `<=> sin^2x \ne cos^2x <=> (1-cos2x)/2 \ne (1+cos2x)/2`

`<=>cos2x \ne 0<=> 2x \ne π/2+kπ <=> x \ne π/4+kπ/2`

3. Hàm số xác định `<=> cos2x \ne 0<=> x \ne π/4+kπ/2 (k \in ZZ)`.

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

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

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

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

c.

\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)

\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)

\(\Leftrightarrow...\)

1.

Chắc đề là \(sin\left[\pi sin2x\right]=1?\)

\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)

Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)

\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)

Thế vào (1)

\(\Rightarrow sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)

Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)

Thế vào (2):

\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\) chắc bạn tự giải tiếp được

3.

\(\Leftrightarrow2sin\left(x+84^0\right).cos\left(60^0\right)=cos20^0\)

\(\Leftrightarrow sin\left(x+84^0\right)=sin70^0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+84^0=70^0+k360^0\\x+84^0=110^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-14^0+k360^0\\x=26^0+k360^0\end{matrix}\right.\)