Cách 2:

 Ta có: $(1-x)(1-y)\ge0 \Leftrightarrow 1+xy\ge x+y$
                                                   $\Leftrightarrow (1+xy)(x+y+2z)\ge(x+y)(x+y+2z)$
Lại có: $z\ge z^2\Rightarrow z(1+xy)\ge z^2$
Cộng 2 BĐT trên lại ta có:
$(1+xy)(x+y+3z)\ge(x+y+z)^2\Rightarrow \dfrac{1}{1+xy}\le\dfrac{x+y+3z}{(x+y+z)^2}$
Tương tự ta có: $\dfrac{1}{1+yz}\le\dfrac{3x+y+z}{(x+y+z)^2};\dfrac{1}{1+xz}\le\dfrac{x+3y+z}{(x+y+z)^2}$
Cộng các BĐT trên lại với nhau ta có đpcm.

Ta có:
$S=\cos4A+2\cos A+2\cos(B+C)\cos(B-C)$
    $=\cos4A+2\cos A[1-\cos(B-C)]$
Vì $\cos A>0,1-\cos(B-C)\ge0 \Rightarrow S\ge\cos4A\ge-1$
Min$S=-1 \Leftrightarrow A=\dfrac{\pi}{4};B=C=\dfrac{3\pi}{8}$

Ta có: $(1-x)(1-y)\ge0 \Leftrightarrow xy+1\ge x+y$
Tương tự: $\left\{\begin{array}{l}yz+1\ge y+z\\xz+1\ge x+z\end{array}\right.$
Khi đó ta có:
     $(x+y+z)\left(\dfrac{1}{xy+1}+\dfrac{1}{yz+1}+\dfrac{1}{zx+1}\right)$
$\le\dfrac{x}{yz+1}+\dfrac{y}{xz+1}+\dfrac{z}{xy+1}+1+1+1$
$\le\dfrac{x}{yz+1}+\dfrac{y}{xz+y}+\dfrac{z}{xy+z}+3$
$=x\left(\dfrac{1}{yz+1}-\dfrac{z}{xz+y}-\dfrac{y}{xy+z}\right)+5$
$\le x\left(1-\dfrac{z}{z+y}-\dfrac{y}{y+z}\right)+5=5$

a.
Ta có:
$\sqrt{x}=0 \Leftrightarrow x=0$
$\sqrt{x}=2-x \Leftrightarrow x=1$
$2-x=0 \Leftrightarrow x=2$
Suy ra:
$S=\int\limits_0^1\sqrt xdx+\int\limits_1^2(2-x)dx$
    $=\dfrac{2}{3}x\sqrt x \left|\begin{array}{l}1\\0\end{array}\right.+\left(2x-\dfrac{x^2}{2}\right)\left|\begin{array}{l}2\\1\end{array}\right.=\dfrac{7}{6}$

a.
Ta có:
$(x-2)^2=4 \Leftrightarrow \left[\begin{array}{l}x=0\\x=4\end{array}\right.$
Suy ra:
$S=\int\limits_0^4|(x-2)^2-4|dx$
     $=\int\limits_0^4(4x-x^2)dx$
     $=\left(2x^2-\dfrac{x^3}{3}\right)\left|\begin{array}{l}4\\0\end{array}\right.=\dfrac{32}{3}$

Ta có:
$4-x^2=2+x^2 \Leftrightarrow 2x^2=2 \Leftrightarrow  x=\pm1$
Suy ra:
$V=\pi\int\limits_{-1}^1(4-x^2)^2dx-\pi\int\limits_{-1}^1(2+x^2)^2dx$
     $=\pi\int\limits_{-1}^1(x^4-8x^2+16)-\pi\int\limits_{-1}^1(x^4+4x^2+4)dx$
     $=\pi\int\limits_{-1}^1(-12x^2+12)dx$
     $=\pi(-4x^3+12x)\left|\begin{array}{l}1\\-1\end{array}\right.=16\pi$

a.
Ta có: $\sqrt x=-x \Leftrightarrow x=0$
Suy ra:
$S(D)=\int\limits_0^5|\sqrt x-(-x)|dx$
           $=\int\limits_0^5(\sqrt x+x)dx$
           $=\left(\dfrac{2}{3}x\sqrt x+\dfrac{x^2}{2}\right)\left|\begin{array}{l}5\\0\end{array}\right.=\dfrac{10}{3}\sqrt5+\dfrac{25}{2}$

b.
Ta có:
$V(D,Oy)=\pi\int\limits_2^4(\sqrt{2y})^2dy$
                    $=2\pi\int\limits_2^4ydy$
                    $=\pi y^2 \left|\begin{array}{l}4\\2\end{array}\right.=12\pi$

a.
Ta có:
$y=\dfrac{x^2}{2}\Leftrightarrow \left[\begin{array}{l} x=\sqrt{2y}\\x=-\sqrt{2y}\end{array} \right.$
Suy ra:
$S(D)=\int\limits_2^4|\sqrt{2y}-(-\sqrt{2y})|dy$
           $=2\sqrt2\int\limits_2^4\sqrt ydy$
           $=\dfrac{4\sqrt2}{3}y\sqrt y \left|\begin{array}{l}4\\2\end{array}\right.=\dfrac{16}{3}(2\sqrt2-1)$

Ta có:
$S=\int\limits_0^1\dfrac{x^2}{8x^3+1}dx$
    $=\dfrac{1}{24}\int\limits_0^1\dfrac{d(8x^3+1)}{8x^3+1}$
    $=\dfrac{1}{24}\ln(8x^3+1)\left|\begin{array}{l}1\\0\end{array}\right.=\dfrac{\ln3}{12}$

Ta có: $\dfrac{x^2}{4}=\dfrac{8}{x^2+4}\Leftrightarrow x^2=4 \Leftrightarrow x=\pm2$
Suy ra diện tích cần tìm là:
$S=\int\limits_{-2}^2\left|\dfrac{x^2}{4}-\dfrac{8}{x^2+4}\right|dx$
    $=\int\limits_{-2}^2\dfrac{8}{x^2+4}dx-\int\limits_{-2}^2\dfrac{x^2}{4}dx$
Đặt: $x=2\tan t \Rightarrow dx=2(\tan^2t+1)dt$
Đổi cận: $x=-2\Rightarrow t=-\dfrac{\pi}{4}$
                $x=2\Rightarrow t=\dfrac{\pi}{4}$
Suy ra:
 $\int\limits_{-2}^2\dfrac{8}{x^2+4}dx=\int\limits_{-\pi/4}^{\pi/4}\dfrac{16(\tan^2t+1)dt}{4\tan^2t+4}$
                          $=\int\limits_{-\pi/4}^{\pi/4}4dt$
                          $=4t \left|\begin{array}{l}\dfrac{\pi}{4}\\\dfrac{-\pi}{4}\end{array}\right.=2\pi$
Mà ta có: $\int\limits_{-2}^2\dfrac{x^2}{4}dx=\dfrac{x^3}{12} \left|\begin{array}{l}2\\-2\end{array}\right.=\dfrac{4}{3}$
Suy ra: $S=2\pi-\dfrac{4}{3}$

Ta có:
$V(D,Ox)=\pi\int\limits_1^2xe^xdx$
                    $=\pi\int\limits_1^2xd(e^x)$
                    $=\pi xe^x \left|\begin{array}{l}2\\1\end{array}\right.-\pi\int\limits_1^2e^xdx$
                    $=2\pi e^2-\pi e-\pi e^x \left|\begin{array}{l}2\\1\end{array}\right.=\pi e^2$

Ta có:
$V(D,Ox)=\pi\int\limits_0^1(1+x^3)^2dx$
                    $=\pi\int\limits_0^1(1+2x^3+x^6)dx$
                    $=\pi\left(x+\dfrac{x^4}{2}+\dfrac{x^7}{7}\right)\left|\begin{array}{l}1\\0\end{array}\right.=\dfrac{23\pi}{14}$

b.
Ta có:
$V(D,Ox)=\pi\int\limits_{-\pi/4}^{\pi/4}\tan^6xdx$
                    $=2\pi\int\limits_0^{\pi/4}\tan^6xdx$
                    $=2\pi\int\limits_0^{\pi/4}(\tan^6x+\tan^4x-\tan^4x)dx$
                    $=2\pi\int\limits_0^{\pi/4}\tan^4x(\tan^2x+1)dx-2\pi\int\limits_0^{\pi/4}(\tan^4x+\tan^2x-\tan^2x)dx$
                    $=2\pi\int\limits_0^{\pi/4}\tan^4xd(\tan x)-2\pi\int\limits_0^{\pi/4}\tan^2x(\tan^2x+1)dx+2\pi\int\limits_0^{\pi/4}(\tan^2x+1-1)dx$
                    $=\dfrac{2\pi\tan^5x}{5}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.-2\pi\int\limits_0^{\pi/4}\tan^2xd(\tan x)+2\pi\int\limits_0^{\pi/4}d(\tan x)-2\pi\int\limits_0^{\pi/4}dx$
                    $=\dfrac{2\pi}{5}-\dfrac{2\pi\tan^3x}{3}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+2\pi\tan x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.-2\pi x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.$
                    $=\dfrac{26\pi}{15}-\dfrac{\pi^2}{2}$