a.
Ta có:
$(x+1)^{2n+1}=C_{2n+1}^{2n+1}x^{2n+1}+C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x+C_{2n+1}^0$
$(x-1)^{2n+1}=C_{2n+1}^{2n+1}x^{2n+1}-C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x-C_{2n+1}^0$
Suy ra:
$(x+1)^{2n+1}-(x-1)^{2n+1}=2C_{2n+1}^{2n}x^{2n}+2C_{2n+1}^{2n-2}x^{2n-2}+\ldots+2C_{2n+1}^2x^2+2C_{2n+1}^0$
Lấy nguyên hàm 2 vế ta được:
$\frac{1}{2n+2}[(x+1)^{2n+2}-(x-1)^{2n+2}]=\frac{2C_{2n+1}^{2n}}{2n+1}x^{2n+1}+2\frac{C_{2n+1}^{2n-2}}{2n-1}x^{2n-1}+\ldots+\frac{2C_{2n+1}^2}{3}x^3+\frac{2C_{2n+1}^0}{1}x$
Cho $x=2$ ta được: $4B=\frac{3^{2n+2}-1}{2n+2}\Rightarrow B=\frac{3^{2n+2}-1}{8(n+1)}$

a.
Ta có:
$(x+1)^{2n+1}=C_{2n+1}^{2n+1}x^{2n+1}+C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x+C_{2n+1}^0$
$\Rightarrow \int\limits_0^t(x+1)^{2n+1}dx=\int\limits_0^t\left(C_{2n+1}^{2n+1}x^{2n+1}+C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x+C_{2n+1}^0\right)dx$
$\Rightarrow \frac{1}{2n+2}(t+1)^{2n+2}=\frac{C_{2n+1}^{2n+1}}{2n+2}t^{2n+2}+\frac{C_{2n+1}^{2n}}{2n+1}t^{2n+1}+\ldots+\frac{C_{2n+1}^1}{2}t^2+\frac{C_{2n+1}^0}{1}t$
$\Rightarrow \int\limits_0^x\frac{1}{2n+2}(t+1)^{2n+2}dt=\int\limits_0^x\left(\frac{C_{2n+1}^{2n+1}}{2n+2}t^{2n+2}+\frac{C_{2n+1}^{2n}}{2n+1}t^{2n+1}+\ldots+\frac{C_{2n+1}^1}{2}t^2+\frac{C_{2n+1}^0}{1}t\right)dt$
$\Rightarrow \frac{1}{(2n+2)(2n+3)}(x+1)^{2n+3}=\frac{C_{2n+1}^{2n+1}}{(2n+2)(2n+3)}x^{2n+3}+\frac{C_{2n+1}^{2n}}{(2n+1)(2n+2)}x^{2n+2}+\ldots+\frac{C_{2n+1}^1}{2.3}x^3+\frac{C_{2n+1}^0}{1.2}x^2$
Cho $x=1$ ta được: $A=\frac{2^{2n+3}}{(2n+2)(2n+3)}$

Ta có:
     $\mathop {\lim }\limits_{x \to 1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$
$=\mathop {\lim }\limits_{x \to 1}\frac{1+x+x^2-3}{1-x^3}$
$=\mathop {\lim }\limits_{x \to 1}\frac{x-1+(x-1)(x+1)}{(1-x)(x^2+x+1)}$
$=\mathop {\lim }\limits_{x \to 1}\frac{-2-x}{x^2+x+1}=-1$

Ta có:
$(x+y+z)(xy+yz+zx)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz$
Suy ra:
$(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz$
                                           $\ge(x+y+z)(xy+yz+zx)-\frac{1}{9}(x+y+z)(xy+yz+zx)$
                                           $=\frac{8}{9}(x+y+z)(xy+yz+zx)$
                                           $\ge\frac{8}{9}(x+y+z).3\sqrt[3]{x^2y^2z^2}=\frac{8}{3}(x+y+z)$
Dấu bằng xảy ra khi: $x=y=z=1$

Ta có:
$\frac{1}{x+1}=2-\frac{1}{y+1}-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{(y+1)(z+1)}}$
$\frac{1}{y+1}=2-\frac{1}{x+1}-\frac{1}{z+1}=\frac{x}{x+1}+\frac{z}{z+1}\ge2\sqrt{\frac{xz}{(x+1)(z+1)}}$
$\frac{1}{z+1}=2-\frac{1}{y+1}-\frac{1}{x+1}=\frac{y}{y+1}+\frac{x}{x+1}\ge2\sqrt{\frac{yz}{(y+1)(x+1)}}$
 Nhân 3 BĐT trên ta có: $xyz\le\frac{1}{8}$
Dấu bằng xảy ra khi: $x=y=z=\frac{1}{2}$

Áp dụng BĐT $(a+b+c)^2\ge 3(ab+bc+ca)$ ta có:
$(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y})^2\geq 3(x^2+y^2+z^2)=9$
$\Rightarrow \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geq 3$
Dấu bằng xảy ra khi: $x=y=z=1$

b.
Ta có: $2x^3+3x^2-1=0 \Leftrightarrow \left[\begin{array}{l}x=-1\\x=\frac{1}{2}\end{array}\right.$
Diện tích hình cần tìm là:
$S=\int\limits_{-1}^{1/2}|2x^3+3x^2-1|dx$
    $=\int\limits_{-1}^{1/2}(1-3x^2-2x^3)dx$
    $=\left(x-x^3-\frac{x^4}{2}\right)\left|\begin{array}{l}\frac{1}{2}\\-1\end{array}\right.=\frac{27}{32}$

a,
Ta có:
$x^2-2x+2=1 \Leftrightarrow x=1$
$x^2+4x+5=1 \Leftrightarrow x=-2$
$x^2-2x+2=x^2+4x+5 \Leftrightarrow x=\frac{-1}{2}$
Diện tích hình cần tìm là:
$S=\int\limits_{-2}^{-1/2}(x^2+4x+5-1)dx+\int\limits_{-1/2}^1(x^2-2x+2-1)dx$
    $=\int\limits_{-2}^{-1/2}(x+2)^2dx+\int\limits_{-1/2}^1(x-1)^2dx$
    $=\frac{(x+2)^3}{3} \left|\begin{array}{l}-\frac{1}{2}\\-2\end{array}\right.+\frac{(x-1)^3}{3} \left|\begin{array}{l}1\\\frac{-1}{2}\end{array}\right.=\frac{9}{4}$

b,
Ta có: $\sqrt x=-x\Leftrightarrow x=0$
Diện tích hình cần tìm là:
$S=\int\limits_0^5(\sqrt x+x)dx$
    $=\frac{2}{3}x\sqrt x \left|\begin{array}{l}5\\0\end{array}\right.+\frac{x^2}{2} \left|\begin{array}{l}5\\0\end{array}\right.$
    $=\frac{10\sqrt5}{3}-\frac{25}{2}$

a,
Ta có:
$y=x^2\ln(1+x^3)=0 \Leftrightarrow x=0$
Diện tích hình cần tìm là:
$S=\int\limits_0^1x^2\ln(1+x^3)dx$
Đặt $1+x^3=t\Rightarrow dt=3x^2dx$
Đổi cận: $x=0\Rightarrow t=1$
                $x=1\Rightarrow t=2$
Ta có:
$S=\frac{1}{3}\int\limits_1^2\ln tdt$
    $=\frac{1}{3}t\ln t \left|\begin{array}{l}2\\1\end{array}\right.-\frac{1}{3}\int\limits_1^2td(\ln t)$
    $=\frac{2}{3}\ln 2-\frac{1}{3}\int\limits_1^2dt$
    $=\frac{1}{3}(2\ln2-1)$

b.
Diện tích hình cần tìm là:
$S=\int\limits_{\pi/6}^{\pi/3}\left|\frac{1}{\sin^2x}-\frac{1}{\cos^2x}\right|dx$
    $=\int\limits_{\pi/6}^{\pi/4}\left(\frac{1}{\sin^2x}-\frac{1}{\cos^2x}\right)dx-\int\limits_{\pi/4}^{\pi/3}\left(\frac{1}{\sin^2x}-\frac{1}{\cos^2x}\right)dx$
    $=(-\cot x-\tan x)\left|\begin{array}{l}\frac{\pi}{4}\\\frac{\pi}{6}\end{array}\right.-(-\cot x-\tan x)\left|\begin{array}{l}\frac{\pi}{3}\\\frac{\pi}{4}\end{array}\right.$
    $=\frac{8}{\sqrt3}-4$

b,
Ta có: $y=x\sqrt{1+x^2}=0\Leftrightarrow x=0$
Diện tích hình cần tìm là:
$S=\int\limits_0^1x\sqrt{1+x^2}dx$
Đặt $1+x^2=t\Rightarrow 2xdx=dt$
Đổi cận: $x=0\Rightarrow t=1$
                $x=1\Rightarrow t=2$
Khi đó, ta có:
$S=\frac{1}{2}\int\limits_0^1\sqrt{1+x^2}.2xdx$
    $=\frac{1}{2}\int\limits_1^2\sqrt tdt$
    $=\frac{1}{3}t\sqrt{t} \left|\begin{array}{l}2\\1\end{array}\right.=\frac{1}{3}(2\sqrt2-1)$

a,
Ta có: $e^x=e^{-x} \Leftrightarrow x=0$
Diện tích hình cần tìm là:
$S=\int\limits_0^1(e^x-e^{-x})dx$
    $=(e^x+e^{-x}) \left|\begin{array}{l}1\\0\end{array}\right.$
    $=e+\frac{1}{e}-2$