a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)

\(\int e^x.\cos xdx\)

= \(\int\cos xd\left(e^x\right)\)

= e^x . cos x - \(\int e^xd\left(\cos x\right)\)

= e^x cos x + \(\int\sin x.e^xdx\)

= e^x cos x + \(\int\sin xd\left(e^x\right)\)

= e^x cos x + sin x . e^x - \(\int e^xd\left(\sin x\right)\)

= e^x ( cos x - sin x ) - \(\int e^x.\cos xdx\)

= \(\int e^x.\cos x=\dfrac{e^x\left(\cos x+\sin x\right)}{2}\)

Vậy a = b = \(\dfrac{1}{2}\)

Lời giải:

Ta có:

\(F(x)=\int f(x)dx=\int e^x\cos xdx\)

Đặt \(\left\{\begin{matrix} u=e^x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^xdx\\ v=\int \cos xdx=\sin x\end{matrix}\right.\)

Do đó:

\(F(x)=\int e^x\cos xdx=e^x\sin x-\int \sin x.e^xdx+c\) (1)

Đặt \(\left\{\begin{matrix} u=e^x\\ dv=\sin xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^xdx\\ v=\int \sin xdx=-cos x\end{matrix}\right.\)

\(\Rightarrow \int \sin x.e^xdx=-\cos x.e^x+\int \cos x.e^xdx+c\) (2)

Từ (1)(2) suy ra:

\(F(x)=e^x.\sin x+\cos x.e^x-\int \cos x.e^xdx+c\)

\(\Leftrightarrow F(x)=e^x\sin x+e^x\cos x-F(x)+c\)

\(\Leftrightarrow F(x)=\frac{1}{2}e^x(\sin x+\cos x)+c\)

Do đó: \(a=b=\frac{1}{2}\)

a: \(y'=4\cdot3x^2-3\cdot2x+2=12x^2-6x+2\)

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

c: \(y'=-2\cdot\left(\sqrt{x}\cdot x\right)'\)

\(=-2\cdot\left(\dfrac{x+x}{2\sqrt{x}}\right)=-2\cdot\dfrac{2x}{2\sqrt{x}}=-2\sqrt{x}\)

d: \(y'=\left(3sinx+4cosx-tanx\right)\)'

\(=3cosx-4sinx+\dfrac{1}{cos^2x}\)

e: \(y'=\left(4^x+2e^x\right)'\)

\(=4^x\cdot ln4+2\cdot e^x\)

f: \(y'=\left(x\cdot lnx\right)'=lnx+1\)

Đáp án C

a) y' = 5cosx -3(-sinx) = 5cosx + 3sinx;

b) = = .

c) y' = cotx +x. = cotx -.

d) + = = (x. cosx -sinx).

e) = = .

f) y' = (√(1+x²))' cos√(1+x²) = cos√(1+x²) = cos√(1+x²).

a.

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

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

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

\(\Leftrightarrow cos^2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

c.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)