Theo giả thiết ta có: $u_3-u_2=12\Leftrightarrow u_3=u_2+12$
Vì $u_1+10;u_2+8;u_3$ lập thành cấp số cộng nên:
$u_1+10+u_3=2(u_2+8)\Leftrightarrow u_1=2u_2+6-u_3\Rightarrow u_1=u_2-6$
Mà $u_n$ là cấp số nhân nên:
$u_1u_3=u_2^2\Leftrightarrow (u_2-6)(u_2+12)=u_2^2\Leftrightarrow u_2=12$
Từ đó suy ra: $u_1=6;u_3=24;u_4=48;u_5=96$
Vậy: $S_5=u_1+u_2+u_3+u_4+u_5=186$

b,
Ta có:
$I=\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin x\cos2xdx$
    $=\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin3xdx-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin xdx$
    $=-\frac{1}{6}\int\limits_{0}^{\frac{\pi}{2}}(x+1)d(\cos3x)+\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}(x+1)d(\cos x)$
    $=-\frac{1}{6}(x+1)\cos3x \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.+\frac{1}{6}\int\limits_{0}^{\frac{\pi}{2}}\cos 3xd(x+1)+\frac{1}{2}(x+1)\cos x \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos xd(x+1)$
    $=\frac{1}{6}-\frac{1}{2}+\frac{1}{6}\int\limits_{0}^{\frac{\pi}{2}}\cos 3xdx-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos xdx$
    $=-\frac{1}{3}+\frac{\sin3x}{18} \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.-\frac{\sin x}{2} \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.$
    $=-\frac{1}{3}-\frac{1}{18}-\frac{1}{2}=-\frac{8}{9}$

a,
Ta có:
$I=\int\limits_{1}^{2} \frac{\ln( x+1)}{x^{2}}dx$
    $=\int\limits_{2}^{1}\ln( x+1)d(\frac{1}{x})$
    $=\frac{\ln(x+1)}{x} \left|\begin{array}{l}1\\2\end{array}\right.-\int\limits_{2}^{1} \frac{d(\ln( x+1))}{x}$
    $=\ln2-\frac{1}{2}\ln3-\int\limits_2^1\frac{dx}{x(x+1)}$
    $=\ln2-\frac{1}{2}\ln3-\int\limits_2^1\left(\frac{1}{x}-\frac{1}{x+1}\right)dx$
    $=\ln2-\frac{1}{2}\ln3-\ln\frac{x}{x+1} \left|\begin{array}{l}1\\2\end{array}\right. $
    $=2\ln2-\frac{1}{2}\ln3+\ln\frac{2}{3}=\ln\frac{8}{3\sqrt3}$

Gọi 4 số cần tìm là $a,b,c,d$.
Theo giải thiết ta có:
$\left\{ \begin{array}{l} a+b+c+d=26&(1)\\a+c=2b&(2)\\bd=c^2&(3)\\b+c=12&(4) \end{array} \right.$
Từ $(4)$ suy ra: $b=12-c$
Từ $(2)$ suy ra: $a=2b-c=24-3c$
Từ $(1)$ suy ra: $d=26-a-b=3c-10$
They vào $(3)$ ta có:
      $(12-c)(3c-10)=c^2$
$\Leftrightarrow 4c^2-46c+120=0$
$\Leftrightarrow c=4$, vì $c\in\mathbb{Z}$
Từ đó suy ra: $a=12,b=8,d=2$

5/
Ta có:
      $4x^2-2(2x+1)+3\le0$
$\Leftrightarrow 4x^2-4x+1\le0$
$\Leftrightarrow (2x-1)^2\le0$
$\Leftrightarrow x=\frac{1}{2}$

3/
Ta có:
     $\frac{1}{x+1}>\frac{1}{x}$
$\Leftrightarrow \frac{1}{x}-\frac{1}{x+1}<0$
$\Leftrightarrow \frac{1}{x(x+1)}<0$
$\Leftrightarrow x(x+1)<0$
$\Leftrightarrow -1<x<0$

1/
Ta có:
$x^2-2x+3=(x-1)^2+2>0, \forall x\in\mathbb{R}$
nên suy ra:
     $(x^2-2x+3)(x^2+4x-5)>0$
$\Leftrightarrow x^2+4x-5>0$
$\Leftrightarrow (x-1)(x+5)>0$
$\Leftrightarrow \left[\begin{array}{l} x>1\\x<-5 \end{array} \right.$

e,
Ta có:
$I=\int\limits_{1}^{2} \frac{dx}{x+x^{5}}$
    $=\int\limits_1^2\left(\frac{1}{x}-\frac{x^3}{1+x^4}\right)dx$
    $=\int\limits_1^2\frac{dx}{x}-\frac{1}{4}\int\limits_1^2\frac{d(x^4+1)}{x^4+1}$
    $=\ln x \left|\begin{array}{l}2\\1\end{array}\right.-\frac{1}{4}\ln(1+x^4)\left|\begin{array}{l}2\\1\end{array}\right. $
    $=\frac{1}{4}\ln\frac{32}{17}$

d,
Ta có:
$I=\int\limits_{1}^{\sqrt{2}}\frac{dx}{2x+x^{3}}$
    $=\int\limits_1^{\sqrt2}\left(\frac{1}{2x}-\frac{x}{2(x^2+2)}\right)dx$
    $=\frac{1}{2}\int\limits_1^{\sqrt2}\frac{dx}{x}-\frac{1}{4}\int\limits_1^{\sqrt2}\frac{d(x^2+2)}{x^2+2}$
    $=\frac{1}{2}\ln x \left|\begin{array}{l}\sqrt2\\1\end{array}\right.-\frac{1}{4}\ln(x^2+2)\left|\begin{array}{l}\sqrt2\\1\end{array}\right. $
    $=\frac{1}{4}\ln\frac{3}{2}$

c,
Ta có:
$I=\int\limits_{0}^{1}\frac{3x^{2}}{x^{2}+4x+4}dx$
    $=\int\limits_0^1\left(3-\frac{12}{x+2}+\frac{12}{x^2+4x+4}\right)dx$
    $=\int\limits_0^13dx-12\int\limits_0^1\frac{d(x+2)}{x+2}+12\int\limits_0^1\frac{d(x+2)}{(x+2)^2}$
    $=3x \left|\begin{array}{l}1\\0\end{array}\right. -12\ln(x+2)\left|\begin{array}{l}1\\0\end{array}\right. -\frac{12}{x+2} \left|\begin{array}{l}1\\0\end{array}\right. $
    $=5+12\ln2-12\ln3$

b,
Ta có:
$I=\int\limits_{-1}^{0}\frac{x}{x^{3}-5x+2}dx$
    $=\int\limits_{-1}^0\left(\frac{2}{7(x-2)}-\frac{2x+1}{7(x^2+2x-1)}\right)dx$
    $=\int\limits_{-1}^0\left(\frac{2}{7(x-2)}-\frac{1+2\sqrt2}{14\sqrt2(x+1+\sqrt2)}+\frac{2\sqrt2-1}{14\sqrt2(x+1-\sqrt2)}\right)dx$
    $=\frac{2}{7}\int\limits_{-1}^0\frac{d(x-2)}{x-2}-\frac{1+2\sqrt2}{14\sqrt2}\int\limits_{-1}^0\frac{d(x+1+\sqrt2)}{x+1+\sqrt2}+\frac{2\sqrt2-1}{14\sqrt2}\int\limits_{-1}^0\frac{d(x+1-\sqrt2)}{x+1-\sqrt2}$
    $=\frac{2}{7}\ln|x-2|\left|\begin{array}{l}0\\-1\end{array}\right.-\frac{1+2\sqrt2}{14\sqrt2}\ln|x+1+\sqrt2|\left|\begin{array}{l}0\\-1\end{array}\right.+\frac{2\sqrt2-1}{14\sqrt2}\ln|x+1-\sqrt2|\left|\begin{array}{l}0\\-1\end{array}\right.$
    $=-\frac{2}{7}\ln3+\frac{3}{7}\ln2-\frac{\sqrt{2}}{14}\ln(1+\sqrt2)$

a,
Ta có:
$I=\int\limits_{0}^{1}\frac{x^{3}-4x^{2}+11x-1}{x^{2}-4x+8}dx$
   $=\int\limits_0^1\left(x+\frac{3(2x-4)}{2(x^2-4x+8)}+\frac{5}{x^2-4x+8}\right)dx$
   $=\int\limits_0^1xdx+\frac{3}{2}\int\limits_0^1\frac{d(x^2-4x+8)}{x^2-4x+8}+5\int\limits_0^1\frac{dx}{(x-2)^2+4}$
   $=\frac{x^2}{2}\left|\begin{array}{l}1\\0\end{array}\right.+\frac{3}{2}\ln(x^2-4x+8)\left|\begin{array}{l}1\\0\end{array}\right.+5\int\limits_0^1\frac{dx}{(x-2)^2+4}$
   $=\frac{1}{2}+\frac{3}{2}\ln\frac{5}{8}+5\int\limits_0^1\frac{dx}{(x-2)^2+4}$
Đặt $x-2=2\tan t, t\in[-\frac{\pi}{2};\frac{\pi}{2}]$
Suy ra: $dx=2d(\tan t)=2(\tan^2t+1)dt$
Đổi cận: $x=0\Rightarrow t=-\frac{\pi}{4}$
                $x=1\Rightarrow t=\arctan(-\frac{1}{2})$
Từ đó suy ra:
$\int\limits_0^1\frac{dx}{(x-2)^2+4}=\int\limits_{-\pi/4}^{\arctan(-1/2)}\frac{2(\tan^2t+1)dt}{4\tan^2t+4}$
                    $=\frac{1}{2}\int\limits_{-\pi/4}^{\arctan(-1/2)}dt$
                    $=\frac{1}{2}t \left|\begin{array}{l}\arctan(-\frac{1}{2})\\-\frac{\pi}{4}\end{array}\right.=\frac{\pi}{8}-\frac{1}{2}\arctan(\frac{1}{2})$
Suy ra: $I=\frac{1}{2}+\frac{3}{2}\ln\frac{5}{8}+\frac{5\pi}{8}-\frac{5}{2}\arctan(\frac{1}{2})$

Bài 2:
Giả sử 4 số cần tìm là: $a;aq;aq^2;aq^3,a\ne0$
Theo đề bài ta có: $\left\{ \begin{array}{l} a+aq+aq^2+aq^3=15\\a^2+a^2q^2+a^2q^4+a^2q^6=85 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} a(1+q+q^2+q^3)=15\\a^2(1+q^2+q^4+q^6)=85 \end{array} \right.$
$\Rightarrow 85(1+q+q^2+q^3)^2=225(1+q^2+q^4+q^6)$
$\Leftrightarrow 14q^6-17q^5-3q^4-34q^3-3q^2-17q+14=0$
$\Leftrightarrow (q-2)(2q-1)(7q^2+q+7)(q^2+1)=0$
$\Leftrightarrow \left[\begin{array}{l} q=2\\q=\frac{1}{2} \end{array} \right.$
Với $q=2\Rightarrow a=\frac{15}{1+q+q^2+q^3}=1$, ta có cấp số nhân: $1;2;4;8$
Với $q=\frac{1}{2}\Rightarrow a=\frac{15}{1+q+q^2+q^3}=8$, ta có cấp số nhân: $8;4;2;1$

Bài 1:
Giả sử công sai của cấp số cộng là $d$.
Ta có:
$u_1=2,u_8=37$
Suy ra: $37=u_1+7d=2+7d\Rightarrow d=5$
Vậy 6 số cần điền là $7;12;17;22;27;32$