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\(\left(sinx+siny\right)^2=3\Leftrightarrow sin^2x+sin^2y+2sinxsiny=3\) (1)

\(\left(cosx-cosy\right)^2=1\Leftrightarrow cos^2x+cos^2y-2cosx.cosy=1\) (2)

Cộng vế với vế (1) và (2):

\(sin^2x+cos^2x+sin^2y+cos^2y-2\left(cosx.cosy-sinx.siny\right)=4\)

\(\Leftrightarrow2-2cos\left(x+y\right)=4\)

\(\Rightarrow cos\left(x+y\right)=-1\)

\(A=\dfrac{cosx+cosy}{cosx-cosy}=\dfrac{2cos\dfrac{x+y}{2}.cos\dfrac{x-y}{2}}{-2sin\dfrac{x+y}{2}.sin\dfrac{x-y}{2}}=-cot\dfrac{x+y}{2}.cot\dfrac{x-y}{2}\)

\(B=\dfrac{sin7x+sin5x}{sin7x-sin5x}=\dfrac{2sin6x.cosx}{2cos6x.sinx}=tan6x.cotx\)

\(x+2y=\dfrac{\pi}{2}\)

\(\Leftrightarrow x+y=\dfrac{\pi}{2}-y\) thay vào A được:

\(A=\dfrac{cos\left(\dfrac{\pi}{2}-y\right)-cosy}{cos\left(\dfrac{\pi}{2}-y\right)+cosy}\)\(=\dfrac{siny-cosy}{siny+cosy}\)\(=\dfrac{\dfrac{\sqrt{2}}{2}.siny-\dfrac{\sqrt{2}}{2}.cosy}{\dfrac{\sqrt{2}}{2}.siny+\dfrac{\sqrt{2}}{2}cosy}\)\(=\dfrac{cos\dfrac{\pi}{4}.siny-sin\dfrac{\pi}{4}.cosy}{sin\dfrac{\pi}{4}.siny+cos\dfrac{\pi}{4}.cosy}\)

\(=\dfrac{sin\left(y-\dfrac{\pi}{4}\right)}{cos\left(y-\dfrac{\pi}{4}\right)}\)\(=tan\left(y-\dfrac{\pi}{4}\right)\)

\(x+2y=\dfrac{\pi}{2}\Rightarrow x+y=\dfrac{\pi}{2}-y\)

\(\Rightarrow cos\left(x+y\right)=cos\left(\dfrac{\pi}{2}-y\right)\)

\(\Rightarrow cos\left(x+y\right)=siny\)

Do đó: \(A=\dfrac{siny-cosy}{siny+cosy}=\dfrac{\sqrt{2}sin\left(y-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(y-\dfrac{\pi}{4}\right)}=tan\left(y-\dfrac{\pi}{4}\right)\)

Lời giải:

Tìm đạo hàm theo biến $y$, bạn chỉ cần coi $x$ là một tham số rồi sử dụng công thức như bình thường thôi.

\(f(y)=y.e^{xy}.\sin y\)

\(\Rightarrow f'(y)=(y.e^{xy})'\sin y+y.e^{xy}(\sin y)'\)

\(=[y'.e^{xy}+y(e^{xy})']\sin y+y.e^{xy}.\cos y\)

\(=(e^{xy}+yxe^{xy})\sin y+y.e^{xy}\cos y\)

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Tính đạo hàm cấp 2.

Theo biến $x$

\(f(x)=e^{xy}\sin y\)

\(\Rightarrow f'(x)=\sin y(e^{xy})'=\sin y.ye^{xy}\)

\(\Rightarrow f''(x)=(y\sin y.e^{xy})'=y\sin y(e^{xy})'=y^2\sin y.e^{xy}\)

Theo biến $y$

\(f(y)=e^{xy}.\sin y\)

\(\Rightarrow f'(y)=(e^{xy})'\sin y+(\sin y)'e^{xy}\)

\(=x.e^{xy}\sin y+\cos y.e^{xy}\)

\(\Rightarrow f''(y)=(xe^{xy}.\sin y+\cos y.e^{xy})'\)

\(=(x.e^{xy}\sin y)'+(\cos y.e^{xy})'\)

\(=(x.e^{xy})'\sin y+(\sin y)'.xe^{xy}+(\cos y)'e^{xy}+\cos y(e^{xy})'\)

\(=x^2e^{xy}.\sin y+\cos y.x.e^{xy}-\sin y.e^{xy}+x\cos y.e^{xy}\)

What a cozy house!

What a cozy house! 

`#ava`

`(1)`What a cozy house!

`(2)` How cozy house is!

`->` Cấu trúc : What + a/ an + adj + noun + S + V!

                        How + adj/adv + S + V/tobe!

\(2sin\left(x+y\right)=sinx+siny\)

\(\Leftrightarrow2.2.sin\dfrac{x+y}{2}.cos\dfrac{x+y}{2}=2.sin\dfrac{x+y}{2}.cos\dfrac{x-y}{2}\)

\(\Leftrightarrow2cos\dfrac{x+y}{2}=cos\dfrac{x-y}{2}\)

\(\Leftrightarrow2\left(cos\dfrac{x}{2}.cos\dfrac{y}{2}-sin\dfrac{x}{2}.sin\dfrac{y}{2}\right)=cos\dfrac{x}{2}.cos\dfrac{y}{2}+sin\dfrac{x}{2}.sin\dfrac{y}{2}\)

\(\Leftrightarrow cos\dfrac{x}{2}.cos\dfrac{y}{2}=3.sin\dfrac{x}{2}.sin\dfrac{y}{2}\)

\(\Leftrightarrow\left(sin\dfrac{x}{2}:cos\dfrac{x}{2}\right).\left(sin\dfrac{y}{2}:cos\dfrac{y}{2}\right)=\dfrac{1}{3}\)

\(\Leftrightarrow tan\dfrac{x}{2}.tan\dfrac{y}{2}=\dfrac{1}{3}\)

⇔2cosx+y2=cosx−y2⇔2cosx+y2=cosx−y2

⇔cosx2.cosy2=3.sinx2.siny2⇔cosx2.cosy2=3.sinx2.siny2

⇔tanx2.tany2=13⇔tanx2.tany2=13