13) y = sin2$x$.sin6x + $sin^{2}8x$
= $\frac{1}{2}.(cos4x - cos8x) + sin^{2}8x$
$\Rightarrow$ y' = $-\frac{1}{2}.sin4x.(4x)' + \frac{1}{2}.sin8x.(8x) + 2.sin8x.(sin8x)'$
= $-\frac{1}{2}.sin4x.4 + \frac{1}{2}.sin8x.8 + 2.sin8x.cos8x.8$
= $-2.sin4x + 4.sin8x + 8.sin16x$

$\lim_{x\rightarrow 0}$ $x$.$sin\frac{2}{x}$

= $\lim_{x\rightarrow 0}$ $2.\frac{x}{2}.sin\frac{2}{x}$

= $\lim_{x\rightarrow 0}$ $2.\frac{sin\frac{2}{x}}{\frac{2}{x}}$ = 2

1) y = 6x² + $\sqrt{x}$ - 8$x^{3}$

$\Rightarrow$ y' = 12x + $\frac{1}{2\sqrt{x}}$ - 24x²

2) y = $\frac{8x^{2}-4x}{2x + 10x^{6}}$

$\Rightarrow$ y' = $\frac{(8x^{2}-4x)'.(2x + 10x^{6})- (8x^{2}-4x).(2x + 10x^{6})'}{(2x + 10x^{6})^{2}}$

= $\frac{(16x-4).(2x + 10x^{6})-(8x^{2}-4x).(2+60x^{5})}{(2x + 10x^{6})^{2}}$

= $\frac{32x^{2}+160x^{7}-8x-40x^{6}-(16x^{2}+480x^{7}-8x-240x^{6})}{(2x + 10x^{6})^{2}}$

= $\frac{-320x^{7}+200x^{6}+16x^{2}}{(2x + 10x^{6})^{2}}$

3) y = 2x + 1 -$\frac{8x}{x^{3}-7}$

$\Rightarrow$ y' = 2 - $\frac{(8x)'.(x^{3}-7)-8x.(x^{3}-7)'}{(x^{3}-7)^{2}}$

= 2 - $\frac{8.(x^{3}-7)-8x.3x^{2}}{(x^{3}-7)^{2}}$

= 2- $\frac{8x^{3}-56-24x^{3}}{(x^{3}-7)^{2}}$ = 2- $\frac{-16x^{3}-56}{(x^{3}-7)^{2}}$

4) y = $x^{8} - \frac{7x^{2}}{2x+5}$

$\Rightarrow$ y' = $8x^{7} - \frac{(7x^{2})'.(2x+5)-7x^{2}.(2x+5)'}{(2x+5)^{2}}$

= $8x^{7} - \frac{14x.(2x+5)-7x^{2}.2}{(2x+5)^{2}}$

= $8x^{7} - \frac{28x^{2}+70x-14x^{2}}{(2x+5)^{2}}$

= $8x^{7} - \frac{14x^{2}+70x}{(2x+5)^{2}}$

8) y = sin3x - $4sin^{3}2x$
$\Rightarrow$ y' = (sin3x)' - ($4sin^{3}2x$)
= 3cos3x - 4.3.sin²2x.(sin2x)'
= 3cos3x - 12.sin²2x.cos2x.(2x)'
= 3cos3x - 12.sin²2x.cos2x.2
= 3cos3x - 12.sin4x.sin2x

$\lim_{x\rightarrow-\infty}$($\sqrt{x^{2}+1}+\sqrt[3]{x^{3}-1}$)

= $\lim_{x\rightarrow-\infty}$$\sqrt{x^{2}+1}+x-x+\sqrt[3]{x^{3}-1}$

= $\lim_{x\rightarrow-\infty}$[($\sqrt{x^{2}+1}+x$)+($\sqrt[3]{x^{3}-1}-x$)]

= $\lim_{x\rightarrow-\infty}$[$\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}-x}$+$\frac{x^{3}-1-x^{3}}{\sqrt[3]{(x^{3}-1)^{2}}+x.\sqrt[3]{x^{3}-1}+x^{2}}$]

= $\lim_{x\rightarrow-\infty}$[$\frac{1}{\sqrt{x^{2}+1}-x}$+$\frac{-1}{\sqrt[3]{(x^{3}-1)^{2}}+x.\sqrt[3]{x^{3}-1}+x^{2}}$]

= $\lim_{x\rightarrow-\infty}$[$\frac{1}{-x.\sqrt{1+\frac{1}{x^{2}}}-x}$+$\frac{-1}{x^{2}.\sqrt[3]{(1-\frac{1}{x^{3}})^{2}}+x^{2}.\sqrt[3]{1-\frac{1}{x^{3}}}+x^{2}}$

= $\lim_{x\rightarrow-\infty}$[$\frac{\frac{1}{-x}}{\sqrt{1+\frac{1}{x^{2}}}+1}$+$\frac{\frac{-1}{x^{2}}}{\sqrt[3]{(1-\frac{1}{x^{3}})^{2}}+\sqrt[3]{(1-\frac{1}{x^{3}})}+1}$]

= 0

$\lim_{n\rightarrow+\infty }$$\frac{\sqrt{n+1}}{\sqrt{n}+1}$

=$\lim_{n\rightarrow+\infty }$$\frac{\sqrt{n}(\sqrt{1+\frac{1}{n}})}{\sqrt{n}(1+\frac{1}{\sqrt{n}})}$

=$\lim_{n\rightarrow+\infty }$$\frac{\sqrt{1+\frac{1}{n}}}{1+\frac{1}{\sqrt{n}}}$ = 1

$\lim_{n\rightarrow+\infty }$$\sqrt{n}$($\sqrt{n+2}-\sqrt{n}$)

=$\lim_{n\rightarrow+\infty }$$\sqrt{n}$($\frac{n+2-n}{\sqrt{n+2}+\sqrt{n}}$)

=$\lim_{n\rightarrow+\infty }$$\frac{2\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}$

=$\lim_{n\rightarrow+\infty }$$\frac{2}{\sqrt{1+\frac{2}{n}}+1}$

=1

a) tu C ke CE$\perp$AC' (1)
BA$\perp$(ACC') $\Rightarrow$ BA$\perp$CE (2)
tu (1) va (2) $\Rightarrow$ CE$\perp$(ABC')
$\Rightarrow$d(C,(ABC'))=CE
$\Rightarrow$CE=$\frac{2a}{\sqrt{5}}$
B'C $\cap$ BC'={K}
$\Rightarrow$ $\frac{d(B',(ABC'))}{d(C,(ABC'))}=\frac{B'K}{KC}=\frac{BB'}{CC'}$=$\frac{a}{2a}=\frac{1}{2}$
$\Rightarrow$ d(B',(ABC'))=$\frac{1}{2}$d(C,(ABC'))=$\frac{1}{2}$CE=$\frac{a\sqrt{5}}{5}$

b) B'C' $\cap$ BC={I}
$\Delta$ABI, ve duong cao BH
$\Delta$BHB', ve duong cao BK
$\Rightarrow$ AI$\perp$(BB'H)
ta co, BK$\subset$(BB'H) $\Rightarrow$ AI$\perp$BK (3)
BK$\perp$HB' (4)
tu (3) va (4) $\Rightarrow$ BK$\perp$(AIC')$\equiv$(AB'C')
$\Rightarrow$ BK=d(B(AB'C'))
M la tung diem BC
$\Rightarrow$ $\frac{d(B,(AB'C'))}{d(M,(AB'C'))}$=$\frac{IB}{IM}$ (5)
$\frac{IB}{IC}=\frac{1}{2}$
$\Rightarrow$ IB=BC=2a
IM=IB+BM=2a+a=3a
(5) $\Rightarrow$ $\frac{d(B,(AB'C'))}{d(M,(AB'C'))}$=$\frac{2a}{3a}$=$\frac{2}{3}$
$\Rightarrow$ d(M,(AB'C'))=$\frac{2}{3}$BK (6)
$\Delta$AIC co B la trung diem AB
$\Rightarrow$ AC² + AI² = 2AB² + $\frac{IC^{2}}{2}$
$\Rightarrow$ AI=a$\sqrt{13}$
$\Rightarrow$ $\widehat{ABI}=135^{\circ}$
$\Rightarrow$ S_$\Delta$ABI=$\frac{1}{2}$BA.BI.sin($\widehat{ABI}$)
$\Rightarrow$ BH=$\frac{a\sqrt{13}}{13}$
$\Delta$BB'H vuong co duong cao BK
$\Rightarrow$ $\frac{1}{BK^{2}}=\frac{1}{BH^{2}}+\frac{1}{BB'^{2}}$
$\Rightarrow$ BK=$\frac{a\sqrt{3}}{4}$
$\Rightarrow$ d(M,(AB'C'))=$\frac{a\sqrt{13}}{8}$

DK : sin $\frac{x}{2}$ $\neq$ 0
$\Leftrightarrow$ x $\neq$ k2$\pi$ (k$\in Z$)

dat t = cot $\frac{x}{2}$ (1)

sin x = $\frac{2t}{1+t^{2}}$ (2)

cos x = $\frac{1-t^{2}}{1+t^{2}}$ (3)
thay (1), (2), (3) vao pt

sin² a + cos² a =1 $\Rightarrow$ cos² a =1 - sin² a = 1- ($\frac{3}{4}$)² = $\frac{7}{16}$
$\Rightarrow$ cosa=$\frac{\sqrt{7}}{4}$ hoac cosa=-$\frac{\sqrt{7}}{4}$
vi sina=$\frac{3}{4}$ $\Rightarrow$ cos a = $\frac{\sqrt{7}}{4}$
tan a = $\frac{sin a}{cos a}$ =......
cot a= $\frac{1}{tan a}$ =.....
tu do thay vao P tinh