\(y'=\dfrac{cosx}{sinx}\), \(y''=-\dfrac{1}{sin^2x}\).
Vì vậy:
\(y'+y''.sinx+tanx=\dfrac{cosx}{sinx}+\dfrac{-1}{sin^2x}.sinx+\dfrac{sinx}{cosx}\)
\(=\dfrac{cosx}{sinx}+\dfrac{-1}{sinx}+\dfrac{sinx}{cosx}\)
\(=\dfrac{cosx-1}{sinx}+\dfrac{sinx}{cosx}\)\(=\dfrac{cos^2x+sin^2x-cosx}{sinx.cosx}=\dfrac{1-cosx}{sinx.cosx}\).
Bạn xem lại đề nhé.

ta có y'=\(-e^{-x}.\sin+e^{-x}.cosx\)

y"=\(e^{-x}.sinx-e^{-x}.cosx-e^{-x}.cosx-e^{-x}.sinx=-2e^{-x.cosx}\)

vậy y"+2y'+2y=\(-2e^{-x}.cosx-2e^{-x}.sinx+2e^{-x}.cosx+2e^{-x}.sinx=0\)

Ta có \(y'=2e^{2x}\sin5x+5e^{2x}\cos5x\)

          \(y"=4e^{2x}\sin5x+10e^{2x}\cos5x+10e^{2x}\cos5x-25e^{2x}\sin5x\)

               \(=-21e^{2x}\sin5x+20e^{2x}\cos5x\)

Vậy \(y"-4y'+29=-21e^{2x}\sin5x+20e^{2x}\cos5x-8e^{2x}\cos5x+29e^{2x}\sin5x=0\)

pt (1) <=>\(x=2+my-4m\) thay vào pt (2) có:

\(\left(2+my-4m\right)m+y=3m+1\)

<=>\(y\left(m^2+1\right)=m+4m^2+1\) (3)

Để hpt có nghiệm <=> pt (3) có nghiệm

<=> \(m^2+1\ne0\) (luôn đúng với mọi m)

=> pt (3) có nghiệm duy nhất => hpt có nghiệm duy nhất với mọi m.

Do x₀,y₀ là 1 nghiệm của hệ => \(\left\{{}\begin{matrix}x_0-my_0=2-4m\\my_0+y_0=3m+1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x_0-2=m\left(y_0-4\right)\\y_0-1=m\left(3-x_0\right)\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}\left(x_0-2\right)\left(3-x_0\right)=m\left(3-x_0\right)\left(y_0-4\right)\\\left(y_0-1\right)\left(y_0-4\right)=m\left(3-x_0\right)\left(y_0-4\right)\end{matrix}\right.\)

=>\(\left(x_0-2\right)\left(3-x_0\right)=\left(y_0-1\right)\left(y_0-4\right)\)

<=>\(5x_0-x_0^2-6=y_0^2-5y_0+4\)

<=>\(x^2_0+y^2_0-5\left(y_0+x_0\right)+10=0\)

Ta có \(y'=\frac{\cos\left(\ln x\right)-\sin\left(\ln x\right)}{x}\)

                 \(\Rightarrow y"=\frac{x.\frac{-\sin\left(\ln x\right)-\cos\left(\ln x\right)}{x}-\left[\cos\left(\ln x\right)-\sin\left(\ln x\right)\right]}{x^2}=\frac{-2\cos\left(\ln x\right)}{x^2}\)

Ta có : 

            \(y+xy'+x^2y"=\sin\left(\ln x\right)+\cos\left(\ln x\right)+\cos\left(\ln x\right)-\sin\left(\ln x\right)-2\cos\left(\ln x\right)=0\)

\(\frac{sin^2a+1}{2.cos^2a}+\frac{1+cos^2a}{2.sin^2a}+1=\frac{tan^2a}{2}+\frac{1}{2cos^2a}+\frac{cot^2a}{2}+\frac{1}{2sin^2a}+1\)

\(=\frac{1}{2}\left(tan^2a+1+tan^2a+cot^2a+1+cot^2a+2\right)\)

\(=\frac{1}{2}\left(2tan^2a+4+2cot^2a\right)=tan^2a+2+cot^2a=\left(tana+cota\right)^2\)

B.

\(\frac{1-4sin^2a.cos^2a}{4sin^2a.cos^2a}=\frac{\frac{1}{cos^4a}-\frac{4sin^2a}{cos^2a}}{\frac{4sin^2a}{cos^2a}}=\frac{\left(\frac{1}{cos^2a}\right)^2-4tan^2a}{4tan^2a}=\frac{\left(1+tan^2a\right)^2-4tan^2a}{4tan^2a}\)

\(=\frac{tan^4a-2tan^2a+1}{4tan^2a}\)

C.

\(\frac{sina+tana}{tana}=\frac{sina}{tana}+1=1+sina.\frac{cosa}{sina}=1+cosa\)

D.

\(tana+\frac{cosa}{1+sina}=\frac{sina}{cosa}+\frac{cosa\left(1-sina\right)}{1-sin^2a}=\frac{sina.cosa}{cos^2a}+\frac{cosa-cosa.sina}{cos^2a}\)

\(=\frac{sina.cosa+cosa-sina.cosa}{cos^2a}=\frac{cosa}{cos^2a}=\frac{1}{cosa}\)

Câu C sai

Ta có \(y'=-a.e^{-x}-2b.e^{-2x};y"=ae^{-x}+4be^{-2x}\)

Vậy \(y"+3y+2y=ae^{-x}+4be^{-2x}-3ae^{-x}-6be^{-2x}+2ae^{-x}+2be^{-2x}=0\)

\(VT=\frac{1-cosx}{sinx}\left[\frac{\left(1+cosx\right)^2}{sin^2x}-1\right]\)

\(=\frac{1-cosx}{sinx}.\left[\frac{2\left(1+cosx\right)-sin^2x}{sin^2x}-1\right]\)

\(=\frac{2\left(1-cos^2x\right)}{sin^3x}-\frac{2\left(1-cosx\right)}{sinx}\)

\(=\frac{2}{sinx}-\frac{2-2cosx}{sinx}\)

\(=\frac{2cosx}{sinx}=2cotx\)