\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)

\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)

\(\Rightarrow P=4\)

\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)

\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)

\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)

giúp mình với . mình đang cần gấp nhé!

\(A=cos^2x.sin^2x\left(sin^4x+cos^4x\right)=\dfrac{1}{4}\left(2sinx.cosx\right)^2\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(=\dfrac{1}{4}sin^22x\left(1-\dfrac{1}{2}sin^22x\right)=\dfrac{1}{8}\left(1-cos4x\right)\left(1-\dfrac{1}{2}\left(1-cos4x\right)\right)\)

\(=\dfrac{1}{8}\left(1-\dfrac{2}{3}\right)\left(1-\dfrac{1}{2}\left(1-\dfrac{2}{3}\right)\right)=\dfrac{5}{144}\)

`x+y=3`

`<=>(x+y)^3=9`

`<=>x^2+2xy+y^2=9`

`<=>2xy+5=9`

`<=>2xy=4`

`<=>xy=2`

`<=>x^2-xy+y^2=3`

`=>M=(x+y)(x^2-xy+y^2)`

`=3.3`

`=9`

\(\Rightarrow\)x²+y²-xy =3

Có M=x³+y³

\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)

Ta có: \(1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha  \ne {90^o})\)

\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)

\( \Leftrightarrow {\cos ^2}\alpha  = \frac{1}{{10}} \Leftrightarrow \cos \alpha  =  \pm \frac{{\sqrt {10} }}{{10}}\)

Vì \({0^o} < \alpha  < {180^o}\) nên \(\sin \alpha  > 0\).

Mà \(\tan \alpha  = 3 > 0 \Rightarrow \cos \alpha  > 0 \Rightarrow \cos \alpha  = \frac{{\sqrt {10} }}{{10}}\)

Lại có: \(\sin \alpha  = \cos \alpha .\tan \alpha  = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)

\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)