Cho biết \(cosx=-\dfrac{1}{2}\)

\(sin^2x+cos^2x=1\Rightarrow sin^2x=1-cos^2x\)

\(\Rightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\)

\(S=4sin^2x+8tan^2x\)

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

\(\Rightarrow S=4\left(\dfrac{3}{4}+2\dfrac{\dfrac{3}{4}}{\dfrac{1}{4}}\right)\)

\(\Rightarrow S=4\left(\dfrac{3}{4}+6\right)\)

\(\Rightarrow S=4.\dfrac{27}{4}=27\)

4.

Gọi H là chân đường cao kẻ từ C xuống đường thẳng d.

Ta có: \(CH=d\left(C;d\right)=\dfrac{\left|-3.2-4.5+4\right|}{\sqrt{3^2+4^2}}=\dfrac{22}{5}\)

Khi đó: \(S_{ABC}=\dfrac{1}{2}CH.AB=\dfrac{1}{2}.\dfrac{22}{5}.AB=15\Rightarrow AB=\dfrac{75}{11}\)

\(\Rightarrow IA=IB=\dfrac{75}{22}\)

Gọi \(A=\left(4m;3m+1\right)\) là điểm cần tìm.

Ta có: \(IA=\dfrac{75}{22}\Leftrightarrow\sqrt{\left(4m-2\right)^2+\left(3m-\dfrac{3}{2}\right)^2}=\dfrac{75}{22}\)

\(\Leftrightarrow\sqrt{25m^2-25m+\dfrac{25}{4}}=\dfrac{75}{22}\)

\(\Leftrightarrow\left|m-\dfrac{1}{2}\right|=\dfrac{15}{22}\)

\(\Leftrightarrow\left[{}\begin{matrix}m-\dfrac{1}{2}=\dfrac{15}{22}\\m-\dfrac{1}{2}=-\dfrac{15}{22}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{13}{11}\\m=-\dfrac{2}{11}\end{matrix}\right.\)

\(m=\dfrac{13}{11}\Rightarrow A=\left(\dfrac{52}{11};\dfrac{50}{11}\right)\Rightarrow B=\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)

Vậy \(A=\left(\dfrac{52}{11};\dfrac{50}{11}\right);B=\left(-\dfrac{8}{11};\dfrac{5}{11}\right)\)

1.

\(P=\left(m;m+1\right)\) là điểm cần tìm 

\(\Rightarrow NP=\sqrt{\left(m-3\right)^2+m^2}=\sqrt{2m^2-6m+9}\)

Ta có: \(NM=NP\)

\(\Leftrightarrow\sqrt{\left(-1-3\right)^2+\left(2-1\right)^2}=\sqrt{2m^2-6m+9}\)

\(\Leftrightarrow m^2-3m-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}P=\left(4;5\right)\\P=\left(-1;0\right)\end{matrix}\right.\)

Vậy \(P=\left(4;5\right)\) hoặc \(P=\left(-1;0\right)\)

2.

\(tanx=-2\Leftrightarrow\dfrac{sinx}{cosx}=-2\Leftrightarrow sinx=-2cosx\)

Khi đó:

\(A=\dfrac{sin^2x+3sinx.cosx-cos^2x+1}{3sin^2x+4sinx.cosx+5cos^2x-2}\)

\(=\dfrac{2sin^2x+3sinx.cosx}{2\left(sin^2x-1\right)+sin^2x+5cos^2x+4sinx.cosx}\)

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

\(=\dfrac{8cos^2x-3.2cos^2x}{4cos^2x+3cos^2x-4.2cos^2x}\)

\(=\dfrac{2cos^2x}{-cos^2x}=-2\)

\(0< =sin^2x< =1\)

=>\(-2< =sin^2x-2< =-1\)

=>\(sin^2x-2< 0\)

\(0< =cos^2x< =1\)

=>\(-2< =cos^2x-2< =-1\)

\(\Leftrightarrow cos^2x-2< 0\)

\(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}\)

\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\cdot\left(1-cos^2x\right)}\)

\(=\sqrt{sin^4x-4sin^xx+4}+\sqrt{cos^4x-4\cdot cos^2x+4}\)

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

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

\(=2-sin^2x+2-cos^2x\)

\(=4-\left(sin^2x+cos^2x\right)=4-1=3\)

\(B=cos^2x+sin^2x+tan^2x\)

\(=1+tan^2x\)

\(=\dfrac{1}{cos^2x}=1:\dfrac{1}{4}=4\)

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

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

\(\Rightarrow P=4\)

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

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

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

Ta có sin²x + cos²x = 1 => sin²x = 1 - cos²x

Do đó P = 3sin²x + cos²x = 3(1 - cos²x) + cos²x

=> P = 3 - 2cos²x

Với cosx = => cos²x = => P= 3 - =