chẳng hiểu gì cả

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

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

\(N=\left(\frac{sinx+\frac{sinx}{cosx}}{cosx+1}\right)^2+1=\left(\frac{sinx.cosx+sinx}{cosx\left(cosx+1\right)}\right)^2+1\)

\(=\left(\frac{sinx\left(cosx+1\right)}{cosx\left(cosx+1\right)}\right)^2+1=tan^2x+1=\frac{1}{cos^2x}\)

\(P=\frac{1-sin^2x.cos^2x}{cos^2x}-cos^2x=\frac{1}{cos^2x}-sin^2x-cos^2x\)

\(=1+tan^2x-\left(sin^2x+cos^2x\right)=1+tan^2x-1=tan^2x\)

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

\(\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}=cosx-sinx\)

Ta có hệ: \(\left\{{}\begin{matrix}3sin^4x-cos^4x=\dfrac{1}{2}\\sin^2x+cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\left(1-cos^2x\right)^2-cos^4x=\dfrac{1}{2}\\sin^2x=1-cos^2x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4cos^4x-12cos^2x+5=0\left(1\right)\\sin^2x=1-cos^2x\left(2\right)\end{matrix}\right.\)

Từ (1) ta có: \(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{5}{2}\left(l\right)\\cos^2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sin^2x=\dfrac{1}{2}\)

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