5:

a: sin x=2*cosx

\(A=\dfrac{6cosx+2cosx-4\cdot8\cdot cos^3x}{cos^3x-2cosx}\)

\(=\dfrac{8-32cos^2x}{cos^2x-2}\)

b: VT=sin^4(pi/2-x)+cos^4(x+pi/2)+6*1/2*sin^22x+1/2*cos4x

=cos^4x+sin^4x+3*sin^2(2x)+1/2*(1-2*sin^2(2x))

=1-2*sin^2x*cos^2x+3*sin^2(2x)+1/2-sin^2(2x)

==3/2=VP

Bán kính hình tròn:
\(18,84:3,14:2=3\left(cm\right)\)
Diện tích hình tròn:
\(3\times3\times3,14=28,26\left(cm^2\right)\)
Đường kính hình tròn:
\(3\times2=6\left(cm\right)\)
Diện tích hình thoi:
\(\dfrac{6\times6}{2}=18\left(cm^2\right)\)
Diện tích phần gạch chéo:
\(28,26-18=10,26\left(cm^2\right)\)

Bài 1:

Vì AD là p/g góc A nên \(\widehat{A_1}=\widehat{A_2}=\dfrac{1}{2}\widehat{BAC}=30^0\)

Mà \(\widehat{A_2}+\widehat{C}+\widehat{D_1}=180^0\Rightarrow\widehat{D_1}=180^0-30^0-40^0=110^0\)

Mà AE//BC nên \(\widehat{EAD}=\widehat{D_1}=110^0\left(so.le.trong\right)\)

Vì DE//AC nên \(\widehat{A_2}=\widehat{D_2}=30^0\left(so.le.trong\right);\widehat{D_3}=\widehat{C}=40^0\left(đồng.vị\right)\)

Vì AE//BC nên \(\widehat{D_3}=\widehat{E}=40^0\)

Vậy các góc tg ADE là \(\widehat{A}=110^0;\widehat{D}=30^0;\widehat{E}=40^0\)

Bài 2:

a, Xét tg ABH và tg DBH có 

\(\left\{{}\begin{matrix}\widehat{BAH}=\widehat{BDH}=90^0\\\widehat{ABH}=\widehat{DBH}\left(BH.là.p/g\right)\\BH.chung\end{matrix}\right.\\ \Rightarrow\Delta ABH=\Delta BDH\left(ch-gn\right)\\ \Rightarrow HA=HD\)

b, Xét tg AHE và tg DHC có

\(\left\{{}\begin{matrix}\widehat{AHE}=\widehat{DHC}\left(đối.đỉnh\right)\\\widehat{HAE}=\widehat{HDC}=90^0\\HA=HD\left(cmt\right)\end{matrix}\right.\\ \Rightarrow\Delta AHE=\Delta DHC\left(g.c.g\right)\\ \Rightarrow CD=AE\)

Mà \(AB=BD\left(\Delta ABH=\Delta DBH\right)\)

\(\Rightarrow AB+AE=BD+CD\\ \Rightarrow BE=BC\)

Vì \(AD=AB\) nên tg ABD cân tại B

Do đó \(\widehat{BAD}=\dfrac{180^0-\widehat{ABC}}{2}\left(1\right)\)

Vì \(BE=BC\) nên tg BEC cân tại B

Do đó \(\widehat{BEC}=\dfrac{180^0-\widehat{ABC}}{2}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\widehat{BAD}=\widehat{BEC}\) mà 2 góc này ở vị trí đồng vị nên AD//EC

4b.

\(\dfrac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{4}{5}\)

\(\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{3}{4}\)

\(tan\left(a+\dfrac{\pi}{3}\right)=\dfrac{tana+tan\left(\dfrac{\pi}{3}\right)}{1-tana.tan\left(\dfrac{\pi}{3}\right)}=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\left(-\dfrac{3}{4}\right).\sqrt{3}}=...\)

c.

\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{5}{13}\)

\(cos\left(\dfrac{\pi}{3}-a\right)=cos\left(\dfrac{\pi}{3}\right).cosa+sin\left(\dfrac{\pi}{3}\right).sina=\dfrac{1}{2}.\dfrac{5}{13}+\left(-\dfrac{12}{13}\right).\dfrac{\sqrt{3}}{2}=...\)

4:

a: -90<a<0

=>cos a>0

cos^2a=1-(-4/5)^2=9/25

=>cosa=3/5

\(sin\left(45-a\right)=sin45\cdot cosa-cos45\cdot sina=\dfrac{\sqrt{2}}{2}\left(cosa-sina\right)\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{3}{5}-\dfrac{4}{5}\right)=\dfrac{-\sqrt{2}}{10}\)

b: pi/2<a<pi

=>cosa<0

cos^2a+sin^2a=0

=>cos^2a=16/25

=>cosa=-4/5

tan a=3/5:(-4/5)=-3/4

\(tan\left(a+\dfrac{pi}{3}\right)=\dfrac{tana+\dfrac{tanpi}{3}}{1-tana\cdot tan\left(\dfrac{pi}{3}\right)}\)

\(=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\dfrac{-3}{4}\cdot\sqrt{3}}=\dfrac{48-25\sqrt{3}}{11}\)

c: 3/2pi<a<pi

=>cosa>0

cos^2a+sin^2a=1

=>cos^2a=25/169

=>cosa=5/13

cos(pi/3-a)

\(=cos\left(\dfrac{pi}{3}\right)\cdot cosa+sin\left(\dfrac{pi}{3}\right)\cdot sina\)

\(=\dfrac{5}{13}\cdot\dfrac{1}{2}+\dfrac{-12}{13}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{5-12\sqrt{3}}{26}\)

\(a,=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\\ b,=x^3-1+x^3-27=2x^3-28\\ =2\left(-1\right)^3-28=-2-28=-30\\ c,=\left[4x-6-4\left(3-x\right)\right]\left[4x-64++4\left(3-x\right)\right]=6\left(8x-18\right)=12\left(4x-9\right)\\ =12\left(-7\cdot4-9\right)=12\left(-37\right)=-444\)

a/

\(\frac{72}{\left(x-2\right)^2}=8\left(x\ne2\right)\Rightarrow\left(x-2\right)^2=9=3^2\)

\(\Rightarrow\left|x-2\right|=3\)

+ Nếu \(x-2\ge0\Rightarrow x\ge2\Rightarrow x-2=3\Rightarrow x=5\) (thoả mãn đk \(x\ge2\) )

+ Nếu \(x-2< 0\Rightarrow x< 2\Rightarrow2-x=3\Rightarrow x=-1\) (Thoả mãn đk \(x< 2\) )

b/

\(75-5\left(x-3\right)^3=700\Rightarrow\left(x-3\right)^3=-125=\left(-5\right)^3\)

\(\Rightarrow x-3=-5\Rightarrow x=-2\)