a.
Ta có:
$S(D)=\int\limits_{-\pi/4}^{\pi/4}|\tan^3x|dx$
           $=2\int\limits_0^{\pi/4}\tan^3xdx$
           $=2\int\limits_0^{\pi/4}(\tan^3x+\tan x-\tan x)dx$
           $=2\int\limits_0^{\pi/4}\tan x(\tan^2x+1)dx-2\int\limits_0^{\pi/4}\dfrac{\sin x}{\cos x}dx$
           $=2\int\limits_0^{\pi/4}\tan xd(\tan x)+2\int\limits_0^{\pi/4}\dfrac{d(\cos x)}{\cos x}$
           $=\tan^2x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+2\ln(\cos x)\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=1-\ln2$

b.
Ta có:
$V(D,Ox)=\pi\int\limits_0^{\pi/2}\sin^2\dfrac{x}{2}\cos^2xdx$
                    $=\dfrac{\pi}{4}\int\limits_0^{\pi/2}(1-\cos x)(1+\cos2x)dx$
                    $=\dfrac{\pi}{4}\int\limits_0^{\pi/2}(1-\cos x+\cos2x-\cos2x\cos x)dx$
                    $=\dfrac{\pi}{4}\int\limits_0^{\pi/2}(1-\cos x+\cos2x-\dfrac{\cos3x+\cos x}{2})dx$
                    $=\dfrac{\pi}{4}\int\limits_0^{\pi/2}(1-\dfrac{3}{2}\cos x+\cos 2x-\dfrac{1}{2}\cos3x)dx$
                    $=\dfrac{\pi}{4}\left(x-\dfrac{3}{2}\sin x+\dfrac{1}{2}\sin2x-\dfrac{1}{6}\sin3x\right)\left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.=\dfrac{\pi^2}{8}-\dfrac{\pi}{3}$

a.
Ta có:
$S(D)=\int\limits_0^{\pi/2}\left|\sin\dfrac{x}{2}\cos x\right|dx$
           $=\dfrac{1}{2}\int\limits_0^{\pi/2}(\sin\dfrac{3x}{2}-\sin\dfrac{x}{2})dx$
           $=\left(-\dfrac{1}{3}\cos\dfrac{3x}{2}+\cos\dfrac{x}{2}\right)\left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.=\dfrac{4}{3}(\sqrt2-1)$

b.
Ta có:
$V(D,Ox)=\pi\int\limits_0^1(-x^2-2x+6)^2dx-\pi\int\limits_0^1(x^2-4x+6)^2dx$
                    $=\pi\int\limits_0^1(x^4+4x^3-8x^2-24x+36)dx-\pi\int_0^1(x^4-8x^3+28x^2-48x+36)dx$
                    $=\pi\int\limits_0^1(12x^3-36x^2+24x)dx$
                    $=\pi(3x^4-12x^3+12x^2)\left|\begin{array}{l}1\\0\end{array}\right.=3\pi$

a.
Ta có:
$x^2-4x+6=-x^2-2x+6\Leftrightarrow 2x^2-2x=0 \Leftrightarrow \left[\begin{array}{l}x=0\\x=1\end{array}\right.$
Suy ra:
$S(D)=\int\limits_0^1|(x^2-4x+6)-(-x^2-2x+6)|dx$
            $=\int\limits_0^1(2x-2x^2)dx$
            $=\left(x^2-\dfrac{2x^3}{3}\right)\left|\begin{array}{l}1\\0\end{array}\right.=\dfrac{1}{3}$

Cách khác:
Ta có:
     $x^6+x^5+x^4+x^3+x^2+x+1$
$=\left(x^3+\dfrac{x^2}{2}+\dfrac{3x}{8}+\dfrac{5}{16}\right)^2+\dfrac{35}{64}\left(x+\dfrac{7}{10}\right)^2+\dfrac{203}{320}>0$
Suy ra phương trình vô nghiệm.

Put $S=a+b,P=ab$
We have,
$A=(a^2+b^2)^2-8a^2b^2$
     $=[(a+b)^2-2ab]^2-8a^2b^2$
     $=(S^2-2P)^2-8P^2$
     $=S^4-4S^2P-4P^2$
     $=-4P^2-2\sqrt2P+\dfrac{1}{2}$
     $=1-4(P+\dfrac{1}{2\sqrt2})^2\le1$
$A$ is a positive integer iff $A=1$ iff $P=-\dfrac{1}{2\sqrt2}$
This implies, $\left\{\begin{array}{l}a+b=\dfrac{\sqrt[4]8}{2}\\ab=-\dfrac{1}{2\sqrt2}\end{array}\right.$
or                     $\left[\begin{array}{l}a=\dfrac{1-\sqrt3}{2\sqrt[4]2};b=\dfrac{1+\sqrt3}{2\sqrt[4]2}\\a=\dfrac{1+\sqrt3}{2\sqrt[4]2};b=\dfrac{1-\sqrt3}{2\sqrt[4]2}\end{array}\right.$

Solution 2:

Applying the AM-GM inequality, we have,
$\left(1+\dfrac{3}{2}x^2\right)+\left(1+\dfrac{3}{2}x^2\right)+1\ge3\sqrt[3]{\left(1+\dfrac{3}{2}x^2\right)^2}$
or $1+x^2\ge\sqrt[3]{\left(1+\dfrac{3}{2}x^2\right)^2}$ or $\sqrt{(1+x^2)^3}\ge1+\dfrac{3}{2}x^2            (1)$
On the other hand,
     $x^4+2x^4+x^2+\dfrac{1}{2}x^2\ge4\sqrt[4]{x^12}=4x^3$
or $3x^4+\dfrac{3}{2}x^2\ge4x^3                           (2)$
From $(1)$ and $(2)$, we get: $3x^4-4x^3\ge1-\sqrt{(1+x^2)^3}$
Equality holds iff $x=0$

Choose $a=\dfrac{-1+\sqrt{8041}}{2}\Rightarrow a^2=a+2010$
Choose $P(x)=(x-a)^{2010}$
We have:
$P(x^2-2010)=(x^2-a-2010)^{2010}$
                            $=(x^2-a^2)^{2010}$
                            $=(x-a)^{2010}(x+a)^{2010}=P(x)(x+a)^{2010}$
It follows that $P(x^2-2010)\,\vdots\,P(x)$, satisfied.

The equation is equivalent to:
      $3x^4-4x^3=(1-\sqrt{1+x^2})(2+x^2+\sqrt{1+x^2})$
or  $x^2(3x^2-4x)=\dfrac{-x^2(2+x^2+\sqrt{1+x^2})}{1+\sqrt{1+x^2}}$
or  $x^2\left(3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\right)=0$
On the other hand, we have:
$3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}=3(x-\dfrac{2}{3})^2+\dfrac{2+5x^2+(\sqrt{1+x^2}-1)^2}{6(1+\sqrt{1+x^2})}>0$
then $x=0$.

Với $k=1$ thì $2^k+3^k=5$, không thỏa mãn.
Với $k\ge2$, ta xét các trường hợp sau:
Nếu $k=2l+1\Rightarrow 2^k+3^k=2^{2l+1}+3.9^l\equiv3( $mod $4)$, loại.
Nếu $k=4l\Rightarrow 2^k+3^k=16^l+81^l\equiv2( $mod $5)$, loại.
Nếu $k=4l+2\Rightarrow 2^k+3^k=4.16^l+9.81^l\equiv3( $mod $5)$, loại.
 Vậy không tồn tại $k$ thỏa mãn.