Với x>2010=>A>0
A=$\frac{x}{2010}$+$\frac{2010}{x-2010}$=$\frac{x^{2}-2010x+2010^{2}}{2010(x-2010)}$
<=>A.2010($x$-2010)=$x^{2}$-2010$x$+$2010^{2}$
<=>$x^{2}$-(2010+2010A)$x$+$2010^{2}$+$2010^{2}$A=0
$\triangle$=$(1005+1005A)^{2}$-$2010^{2}$-$2010^{2}$A $\geq$0
<=>$A^{2}$-2A-3$\geq$0
<=>A$\geq$3 (vì A>0)
Vay MinA=3<=>$x^{2}$-8040$x$-16160400=0
                  <=>$x$=4020(nhan)

b.$x^{4}$+$\sqrt{x^{2}+2005}$=2005
<=>$x^{4}$+$x^{2}$+$\frac{1}{4}$=$x^{2}$+2005-$\sqrt{x^{2}+2005}$+$\frac{1}{4}$
<=>$(x^{2}+\frac{1}{2})^{2}$-$(\sqrt{x^{2}+2005}-\frac{1}{2})^{2}$=0
<=>($x^{2}$+$\sqrt{x^{2}+2005}$)($x^{2}$-$\sqrt{x^{2}+2005}$+1)=0
<=>$x^{2}$-$\sqrt{x^{2}+2005}$+1=0($x^{2}$+$\sqrt{x^{2}+2005}$$\ne$0)
<=>$x^{2}$+2005-$\sqrt{x^{2}+2005}$+$\frac{1}{4}$=$\frac{8017}{4}$
<=>$(\sqrt{x^{2}+2005}-\frac{1}{2})^{2}$=$\frac{8017}{4}$
<=>$\sqrt{x^{2}+2005}$=$\pm$$\frac{\sqrt{8017}}{2}$+$\frac{1}{2}$
<=>$\sqrt{x^{2}+2005}$=$\frac{\sqrt{8017}}{2}$+$\frac{1}{2}$ ($\sqrt{x^{2}+2005}$>0)
<=>$x^{2}$=$\frac{-1+\sqrt{8017}}{2}$
<=>$x$=$\pm$$\sqrt{\frac{-1+\sqrt{8017}}{2}}$

$\sqrt[3]{x^{2}+1}$=$\sqrt{x+1}$ (1)
dk: x$\geq$-1
Đặt b=$\sqrt[3]{x^{2}+1}$<=>$b^{3}$=$x^{2}$+1
a=$\sqrt{x+1}$ (a$\geq$0) <=>$a^{4}$=$x^{2}$+ 2$x$+1=$b^{3}$+2$a^{2}$-2 (2)
Từ (1) ta có a=b (3) =>b$\geq$0
The (3) vào (2) ta được $a^{4}$-$a^{3}$-2$a^{2}$+2=0
<=>$a^{3}$(a-1)-2($a^{2}$-1)=0
<=>(a-1)($a^{3}$-2a-2)=0
<=>a=1(nhan) v $a^{3}$-2a-2=0
.a=1<=>$\sqrt{x+1}$=1<=>x=0(nhan)
.$a^{3}$-2a-2=0 (4)
Đặt a=u+v
Pt (4) trở thành
$(u+v)^{3}$-2(u+v)-2=0
<=>$u^{3}$+$v^{3}$+3uv(u+v)-2(u+v)-2=0
<=>$u^{3}$+$v^{3}$+(3uv-2)(u+v)-2=0
Chọn uv=$\frac{2}{3}$<=>$u^{3}$$v^{3}$=$\frac{8}{27}$
uv=$\frac{2}{3}$=>$u^{3}$+$v^{3}$=2
$u^{3}$ và $v^{3}$ là nghiệm của pt
$X^{2}$-2$X$+$\frac{8}{27}$=0
<=>$X$=$\frac{9\pm\sqrt{57}}{9}$
=>a=$\sqrt[3]{\frac{9+\sqrt{57}}{9}}$+$\sqrt[3]{\frac{9-\sqrt{57}}{9}}$ (nhan)
<=>$\sqrt{x+1}$=$\sqrt[3]{\frac{9+\sqrt{57}}{9}}$+$\sqrt[3]{\frac{9-\sqrt{57}}{9}}$
<=>$x$=$(\sqrt[3]{\frac{9+\sqrt{57}}{9}}$+$\sqrt[3]{\frac{9-\sqrt{57}}{9}})^{2}$-1(nhan)
Vậy S={0;$(\sqrt[3]{\frac{9+\sqrt{57}}{9}}$+$\sqrt[3]{\frac{9-\sqrt{57}}{9}})^{2}$-1}

1.a.
dk:x>0
y=$\frac{x^{2}+\sqrt{x}}{x-\sqrt{x}+1}$+1-$\frac{2x+\sqrt{x}}{\sqrt{x}}$=$\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}$+1-$\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}$
=x+$\sqrt{x}$+1-2$\sqrt{x}$-1=x-$\sqrt{x}$
để y=2$\Leftrightarrow$x-$\sqrt{x}$=2$\Leftrightarrow$$\sqrt{x}$=2($\sqrt{x}$>0)$\Leftrightarrow$x=4(n)
c.
y=x-$\sqrt{x}$+$\frac{1}{4}$-$\frac{1}{4}$=$(\sqrt{x}-\frac{1}{2})^{2}$-$\frac{1}{4}$$\geq$-$\frac{1}{4}$
Vậy Miny=-$\frac{1}{4}$$\Leftrightarrow$x=$\frac{1}{4}$(n)
2.
dk:x$\neq$0..áp dụng bdt Cauchy
M=$\frac{x^{4}+4x^{2}+1}{x^{2}}$=$x^{2}$+4+$\frac{1}{x^{2}}$$\geq$2+4=6
Vậy MinM=6$\Leftrightarrow$x=$\pm$1

Cho các số dương a;b;c thỏa mãn $a+b+c\leq$3. Chứng minh rằng:
$\frac{1}{a^{2}+b^{2}+c^{2}}+\frac{2009}{ab+bc+ca}\geq670$

$\begin{cases} x^{5} + y^{5} + z^{5}=3 (1)\\ x^{6} + y^{6} + z^{6}=3(2) \end{cases}$
Từ pt (2) <=> $x^{6}$=3-$y^{6}$-$z^{6}$$\leq $3
<=>$\left| {x} \right|$$\leq$ $\sqrt[6]{3}$
Tương tự $\left| {y} \right|$$\leq$$\sqrt[6]{3}$,$\left| {z} \right|$$\leq$$\sqrt[6]{3}$
Ta có$x^{6}$$\geq$$x^{5}$,$y^{6}$$\geq$$y^{5}$,$z^{6}$$\geq$$z^{5}$
=>$x^{6}$+$y^{6}$+$z^{6}$$\geq$$x^{5}$+$y^{5}$+$z^{5}$
Mà theo hpt $x^{6}$+$y^{6}$+$z^{6}$=$x^{5}$+$y^{5}$+$z^{5}$
Dấu''='' xảy ra <=>$\begin{cases}x^{6}=x^{5}\\y^{6}=y^{5} \\ z^{6}=z^{5} \end{cases}$
Do đó $\begin{cases} x^{5} + y^{5} + z^{5}=3 \\ x^{6} + y^{6} + z^{6}=3\end{cases}$ =>$\begin{cases}x^{6}=x^{5}\\y^{6}=y^{5} \\ z^{6}=z^{5} \end{cases}$ <=>$\begin{cases}x=0;1 \\ y=0;1\\z=0;1 \end{cases}$
The vao hpt ta nhận $\begin{cases}x=1 \\ y=1\\z=1 \end{cases}$

$x^{2}$ + 4$x$ - $y^{2}$ + 4 = ($x^{2}$ + 4$x$ +4) - $y^{2}$ = $(x+2)^{2}$ - $y^{2} = (x+y+2)(x-y+2)$