a) Gọi $SG_1, SG_2$ cắt $BC,AD$ tại $R, F$ thì $R,F$ là trung điểm của $BC,AD$.
 Ta có 
 $SG_1 : SR = SG_2 : SF =2 : 3 \Rightarrow G_1G_2 \parallel RF \parallel  AB \parallel  CD \Rightarrow $ đpcm.

Vì $G_1, G_2$ là các trọng tâm nên ta có
$CG_1 : CM = CG_2 : CP =2:3 \Rightarrow G_1G_2 \parallel MP \Rightarrow G_1G_2 \parallel SB \Rightarrow G_1G_2 \parallel (SBC)$

$MN \parallel EF \Rightarrow \begin{cases}MN \parallel (ABC) \\ MN \parallel (BCD) \end{cases}$

Gọi $J$ là trung điểm $AD$. Do $G$ là trọng tâm của $\triangle ABD$ nên $BG=2BJ$.
Mà $BI=2IC$ nên $IG \parallel CJ \Rightarrow IG \parallel (ACD)$

Bằng phương pháp đồ thị ta thấy rằng PT trên chỉ có ba nghiệm.
Kiểm tra rằng $f(0)=0, f(0,5)f(1) <0, f(-0,5)f(-1) <0$ như vậy có ba nghiệm thỏa mãn
$x_1=0, x_2 \in (0,5;1), x_3 \in (-1;-0,5)$

b) Ta có
$ \frac{\sin2x}{\sin^{2}x + 2\cos^{2}x}= \frac{\sin2x}{1+\cos^{2}x}= -\frac{(1+\cos^2x)'}{1+\cos^{2}x}$
Suy ra
$\int\limits_{0}^{\pi/4} \frac{\sin2x}{1+\cos^{2}x}dx=\int\limits_{0}^{\pi/4} -\frac{(1+\cos^2x)'}{1+\cos^{2}x}dx=\left[ {-\ln(1+\cos^{2}x)} \right]_{0}^{\pi/4}=\ln\frac{4}{3}$

a)
$I=\int\limits_{-\pi}^{\pi}\sin mx\sin nxdx$
 $=\int\limits_{-\pi}^{\pi} \frac{1}{2}(-\cos(m+n)x+\cos(m-n)x)dx$
+ Nếu $m=n$
$I=\int\limits_{-\pi}^{\pi} \frac{1}{2}(-\cos2mx+1)dx=\left[ {- \frac{1}{4}\sin2mx+\frac{1}{2}x} \right]_{-\pi}^{\pi}=\pi$
+ Nếu $m=-n$
$I=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos2mx-1)dx=\left[ { \frac{1}{4}\cos2mx-\frac{1}{2}x} \right]_{-\pi}^{\pi}=-\pi$
+ Nếu $m\ne \pm n$
 $I=\frac{1}{2}(\frac{-\sin(m+n)x}{m+n}+\frac{\sin(m-n)x}{m-n})dx\left|\begin{array}{l}\pi\\-\pi\end{array}\right.=\frac{\sin(m+n)\pi}{m+n}+\frac{\sin(m-n)\pi}{m-n}$

Ta có
$\dfrac{\cos x \ln(\sin x)}{\sin^{2}x}=-\dfrac{\cos x -\cos x( \ln(\sin x)+1)}{\sin^{2}x}=-\dfrac{\sin x.( \ln(\sin x)+1)' -(\sin x)'( \ln(\sin x)+1)}{\sin^{2}x}$
suy ra
$ \int\limits_{\pi/4}^{\pi/2} \dfrac{\cos x.\ln(\sin x)dx}{\sin^{2}x} =\left[ {-\dfrac{ \ln(\sin x)+1}{\sin x}} \right]_{\pi/4}^{\pi/2}=-1-\dfrac{\ln 2-2}{\sqrt 2}$

b) Ta có
$\int\limits_{0}^{\pi/2}\frac{\sin 2x }{4-\cos ^{2}x}dx=\int\limits_{0}^{\pi/2}\frac{-2\sin 2x }{\cos 2x -7}dx=\int\limits_{0}^{\pi/2}\frac{(\cos 2x-7)'}{\cos 2x -7}dx=\left[ {\ln|\cos 2x -7|} \right]_{0}^{\pi/2}= \ln\dfrac{4}{3} $

a)
$\dfrac{1}{\tan x+1}=\dfrac{\cos x}{\sin x +\cos x}=\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{\cos x-\sin x}{\sin x +\cos x}=\dfrac{1}{2}(x)'+ \dfrac{1}{2}(\ln |\sin x+ \cos x|)'$
suy ra
$\int\limits_{0}^{\pi/4}\dfrac{1}{\tan x+1}dx=\dfrac{1}{2}\left[ {x+\ln |\sin x+ \cos x|} \right]_{0}^{\pi/4}=\dfrac{\pi+\ln 4}{8}$

b)
$ \int \tan^4x dx = \int(\frac{1}{\cos^2 x}\tan^2 x - \tan^2 x) dx = \frac{\tan^3 x}{3} + \int \frac{\cos^2 x - 1}{\cos^2 x} dx= \frac{\tan^3 x}{3} + x - \tan x +C$
Suy ra 
 $ \int\limits_{\pi/4}^{\pi/3} \tan^{4}xdx=\left[ { \frac{\tan^3 x}{3} + x - \tan x } \right]_{\pi/4}^{\pi/3}=\dfrac{\pi+8}{12}$

b)

b)
$I=\int\limits_{-\pi}^{\pi}\cos mx\cos nxdx$
 $=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos(m+n)x+\cos(m-n)x)dx$
+ Nếu $m=n$
$I=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos2mx+1)dx=\left[ { \frac{1}{4}\sin2mx+\frac{1}{2}x} \right]_{-\pi}^{\pi}=\pi$
+ Nếu $m=-n$
$I=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos2mx+1)dx=\left[ { \frac{1}{4}\sin2mx+\frac{1}{2}x} \right]_{-\pi}^{\pi}=\pi$
+ Nếu $m\ne \pm n$
 $I=\frac{1}{2}(\frac{\sin(m+n)x}{m+n}+\frac{\sin(m-n)x}{m-n})dx\left|\begin{array}{l}\pi\\-\pi\end{array}\right.=\frac{\sin(m+n)\pi}{m+n}+\frac{\sin(m-n)\pi}{m-n}$

a) $ \int\limits_{0}^{\pi/2} (2\sin x+3)\cos xdx= \int\limits_{0}^{\pi/2} 2\sin x cos xdx+ 3\int\limits_{0}^{\pi/2}\cos xdx$
$=\left[ {\sin^2 x +3\sin x} \right]_{0}^{\pi/2}=4$

b) Ta có
$2\sin6x.\sin2x=\cos 4x -\cos 8x$
Suy ra
$ \int\limits_{0}^{\pi/6} (\sin6x.\sin2x-6)dx= \left[ {-6x +\dfrac{1}{8}\sin 4x -\dfrac{1}{16}\sin 8x} \right]_{0}^{\pi/6}=\dfrac{3\sqrt 3}{32}-\pi$