tôi nói

a) Ta có (\sin x+\cos x)^{2}=\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x=1+2 \sin x \cos x(sinx+cosx)2=sin2x+2sinxcosx+cos2x=1+2sinxcosx (*)
Mặt khác \sin x+\cos x=msinx+cosx=m nên m^{2}=1+2 \sin \alpha \cos \alpham2=1+2sinαcosα hay \sin \alpha \cos \alpha=\dfrac{m^{2}-1}{2}sinαcosα=2m2−1​
Đặt A=\left|\sin ^{4} x-\cos ^{4} x\right|A=∣∣​sin4x−cos4x∣∣​. Ta có
A=\left|\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)\right|=|(\sin x+\cos x)(\sin x-\cos x)|A=∣∣​(sin2x+cos2x)(sin2x−cos2x)∣∣​=∣(sinx+cosx)(sinx−cosx)∣
\Rightarrow A^{2}=(\sin x+\cos x)^{2}(\sin x-\cos x)^{2}=(1+2 \sin x \cos x)(1-2 \sin x \cos x)⇒A2=(sinx+cosx)2(sinx−cosx)2=(1+2sinxcosx)(1−2sinxcosx)

\Rightarrow A^{2}=\left(1+\dfrac{m^{2}-1}{2}\right)\left(1-\dfrac{m^{2}-1}{2}\right)=\dfrac{3+2 m^{2}-m^{4}}{4}⇒A2=(1+2m2−1​)(1−2m2−1​)=43+2m2−m4​
Vậy A=\dfrac{\sqrt{3+2 m^{2}-m^{4}}}{2}A=23+2m2−m4​​

b) Ta có 2 \sin x \cos x \leq \sin ^{2} x+\cos ^{2} x=12sinxcosx≤sin2x+cos2x=1 kết hợp với (*)(∗) suy ra

(\sin x+\cos x)^{2} \leq 2 \Rightarrow|\sin x+\cos x| \leq \sqrt{2}(sinx+cosx)2≤2⇒∣sinx+cosx∣≤2​

Vậy |m| \leq \sqrt{2}∣m∣≤2​.

​

a) Ta có $A=\genfrac{}{}{0ex}{}{\tan \alpha +3\genfrac{}{}{0ex}{}{1}{\tan \alpha }}{\tan \alpha +\genfrac{}{}{0ex}{}{1}{\tan \alpha }}=\genfrac{}{}{0ex}{}{{\tan }^2\alpha +3}{{\tan }^2\alpha +1}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{1}{{\cos }^2\alpha }+2}{\genfrac{}{}{0ex}{}{1}{{\cos }^2\alpha }}=1+2{\cos }^2\alpha$ Suy ra $A=1+2\cdot \genfrac{}{}{0ex}{}{9}{16}=\genfrac{}{}{0ex}{}{17}{8}$.

b) $B=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{\sin \alpha }{{\cos }^3\alpha }-\genfrac{}{}{0ex}{}{\cos \alpha }{{\cos }^3\alpha }}{\genfrac{}{}{0ex}{}{{\sin }^3\alpha }{{\cos }^3\alpha }+\genfrac{}{}{0ex}{}{3{\cos }^3\alpha }{{\cos }^3\alpha }+\genfrac{}{}{0ex}{}{2\sin \alpha }{{\cos }^3\alpha }}=\genfrac{}{}{0ex}{}{\tan \alpha \left({\tan }^2\alpha +1\right)-\left({\tan }^2\alpha +1\right)}{{\tan }^3\alpha +3+2\tan \alpha \left({\tan }^2\alpha +1\right)}$.

Suy ra $B=\genfrac{}{}{0ex}{}{\sqrt{2}(2+1)-(2+1)}{2\sqrt{2}+3+2\sqrt{2}(2+1)}=\genfrac{}{}{0ex}{}{3(\sqrt{2}-1)}{3+8\sqrt{2}}$.

a) Vì $90^{\circ }<\alpha <180^{\circ }$ nên $\cos \alpha <0$ mặt khác ${\sin }^2\alpha +{\cos }^2\alpha =1$ suy ra $\cos \alpha =-\sqrt{1-{\sin }^2\alpha }=-\sqrt{1-\genfrac{}{}{0ex}{}{1}{9}}=-\genfrac{}{}{0ex}{}{2\sqrt{2}}{3}$.

Do đó $\tan \alpha =\genfrac{}{}{0ex}{}{\sin \alpha }{\cos \alpha }=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{1}{3}}{-\genfrac{}{}{0ex}{}{2\sqrt{2}}{3}}=-\genfrac{}{}{0ex}{}{1}{2\sqrt{2}}$.

b) Vì ${\sin }^2\alpha +{\cos }^2\alpha =1$ nên $\sin \alpha =\sqrt{1-{\cos }^2\alpha }=\sqrt{1-\genfrac{}{}{0ex}{}{4}{9}}=\genfrac{}{}{0ex}{}{\sqrt{5}}{3}$ và $\cot \alpha =\genfrac{}{}{0ex}{}{\cos \alpha }{\sin \alpha }=\genfrac{}{}{0ex}{}{-\genfrac{}{}{0ex}{}{2}{3}}{\genfrac{}{}{0ex}{}{\sqrt{5}}{3}}=-\genfrac{}{}{0ex}{}{2}{\sqrt{5}}$.

c) Vì $\tan \gamma =-2\sqrt{2}<0\Rightarrow \cos \alpha <0$ mặt khác ${\tan }^2\alpha +1=\genfrac{}{}{0ex}{}{1}{{\cos }^2\alpha }$ nên $\cos \alpha =-\sqrt{\genfrac{}{}{0ex}{}{1}{{\tan }^2+1}}=-\sqrt{\genfrac{}{}{0ex}{}{1}{8+1}}=-\genfrac{}{}{0ex}{}{1}{3}$.
Ta có $\tan \alpha =\genfrac{}{}{0ex}{}{\sin \alpha }{\cos \alpha }\Rightarrow \sin \alpha =\tan \alpha \cdot \cos \alpha =-2\sqrt{2}\cdot \left(-\genfrac{}{}{0ex}{}{1}{3}\right)=\genfrac{}{}{0ex}{}{2\sqrt{2}}{3}$ $\Rightarrow \cot \alpha =\genfrac{}{}{0ex}{}{\cos \alpha }{\sin \alpha }=\genfrac{}{}{0ex}{}{-\genfrac{}{}{0ex}{}{1}{3}}{\genfrac{}{}{0ex}{}{2\sqrt{2}}{3}}=-\genfrac{}{}{0ex}{}{1}{2\sqrt{2}}$.

Tìm hiệu của số tròn chục lớn nhất có 2 chữ số 

$\begin{array}{rl}P=& \sqrt{{\left(1-{\cos }^{2}x\right)}^2+6{\cos }^{2}x+3{\cos }^{4}x}+\sqrt{{\left(1-{\sin }^{2}x\right)}^2+6{\sin }^{2}x+3{\sin }^{4}x}\\ & =\sqrt{4{\cos }^{4}x+4{\cos }^{2}x+1}+\sqrt{4{\sin }^{4}x+4{\sin }^{2}x+1}\\ & =\sqrt{{\left(2{\cos }^{2}x+1\right)}^2}+\sqrt{{\left(2{\sin }^{2}x+1\right)}^2}\\ & =2{\cos }^{2}x+1+2{\sin }^{2}x+1\\ & =3\end{array}$

Vậy $P$ không phụ thuộc vào $x$.

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