Ta có : \(\left(SBC\right)\cap\left(ABC\right)=BC\)

Lấy H là TĐ của BC \(\Rightarrow AH\perp BC\)

SA \(\perp\left(ABC\right)\Rightarrow SA\perp AB;AC\) 

\(\Delta SAB;\Delta SAC\perp\) tại A  có : \(SB=\sqrt{SA^2+AB^2}=\sqrt{SA^2+AC^2}=SC\)

\(\Rightarrow\Delta SBC\) cân tại S . Suy ra : \(SH\perp BC\)

Suy ra : \(\left(\left(SBC\right);\left(ABC\right)\right)=\left(HA;HS\right)=\widehat{SHA}\)

Tính được : AH = \(\dfrac{a\sqrt{3}}{2}\)

\(\Delta SAH\) vuông tại A có : \(tan\widehat{SHA}=\dfrac{SA}{HA}=\dfrac{a\sqrt{3}}{2}:\dfrac{a\sqrt{3}}{2}=1\Rightarrow\widehat{SHA}=45^o\)

Vậy ... 

A luôn có nghĩa vì : \(x^2-3x+4>0;5x^2-2x+1>0\)

\(\left(...\right)=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2}{\sqrt{2x+7}+3}=\dfrac{1}{3}\)

\(\lim\limits_{x\rightarrow a}\dfrac{x^4-a^4}{x^2-a^2}=\lim\limits_{x\rightarrow a}\left(x^2+a^2\right)=2a^2\)

\(\left(...\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{3-\dfrac{5}{x}+\dfrac{2}{x^2}}{2+\dfrac{4}{x}-\dfrac{1}{x^2}}=\dfrac{3}{2}\)

\(\left(tan3x\right)'=\dfrac{3}{cos^23x}\)  ; \(\left(cot\left(x^2\right)\right)'=-\dfrac{2x}{sin^2x^2}\)  

\(\left[sin\left(2x-\dfrac{\pi}{3}\right)\right]'=2cos\left(2x-\dfrac{\pi}{3}\right)\)

Dễ thấy : \(f'\left(-2\right)=5\) . G/s đths có p/t tiếp tuyến tại \(M\left(-2;y_M\right)\)

Suy ra : \(y_M=5.\left(-2\right)-3=-13=f\left(-2\right)\)

Ta có : \(lim\dfrac{an^3+bn^2+2n+4}{n^2+1}=lim\dfrac{an+b+\dfrac{2}{n}+\dfrac{4}{n^2}}{1+\dfrac{1}{n^2}}=1\)  \(\Rightarrow a=0\)

Với a = 0 ; \(lim\dfrac{b+\dfrac{2}{n}+\dfrac{4}{n^2}}{1+\dfrac{1}{n^2}}=1\Rightarrow b=1\)  Vậy ... 

Ta có : \(D=cos^2\left(x+y\right)+cos^2\left(x-y\right)-cos2x.cos2y\)

\(=\left[cosx.cosy-sinx.siny\right]^2+\left[cosx.cosy+sinx.siny\right]^2-cos2x.cos2y\)

\(=2\left[cos^2x.cos^2y+sin^2x.sin^2y\right]-\left(cos^2x-sin^2x\right)\left(cos^2y-sin^2y\right)\)

\(=cos^2x.cos^2y+sin^2x.sin^2y+sin^2x.cos^2y+cos^2x.sin^2y\)

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