\((x+y)^3-(x-y)^3\)

\(=x^3+3x^2y+3xy^2+y^3-(x^3-3x^2y+3xy^2-y^3)\)

\(=6x^2y+2y^3\)

`a)sqrt{(sqrt7-4)^2}+sqrt7`

`=|sqrt7-4|+sqrt7`

`=4-sqrt7+sqrt7=4`

`b)\sqrt{81a}-sqrt{144a}+sqrt{36a}(a>=0)`

`=9sqrta-12sqrta+6sqrta=3sqrta`

`(1-5x)/(x-1)>=1`

`<=>(1-5x)/(x-1)-1>=0`

`<=>(1-5x-x+1)/(x-1)>=0`

`<=>(2-6x)/(x-1)>=0(x ne 1)`

`<=>(6x-2)/(x-1)<=0`

`<=>(x-2/6)/(x-1)<=0`

`TH1:x-2/6>=0,x-1<0` 

`<=>2/6<=x<1(TM)`

`TH2:x-2/6<=0,x-1>0`

`<=>1<x<=2/6`(vô lý)

Vậy `2/6<=x<1`

B sai rồi bạn.

$a)ĐK:8x+2\ge 0$

$\to 8x \ge -2$

$\to x \ge -\dfrac14$

$b)ĐK:\dfrac{-5}{6-3x} \ge 0(x \ne 2)$

Mà $-5<0$

$\to 6-3x<0$

$\to 6<3x$

$\to x>2$

$*A=x-2\sqrt{x-2}+3(x \ge 2)$

$=x-2-2\sqrt{x-2}+1+4$

$=(\sqrt{x-2}-1)^2+4 \ge 4$

Dấu "=" xảy ra khi $\sqrt{x-2}-1=0 \Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$

Đặt `A(x)=0`

`<=>4x-2(3x-5)+2=0`

`<=>4x-6x+10+2=0`

`<=>12-2x=0`

`<=>12=2x`

`<=>x=6`

Vậy x=6 là nghiệm A(x)

`A=\sqrt{1+2008^2+2008^2/2009^2}+2008/2009`

`=\sqrt{1+2008^2+2.2008+2008^2/2009^2-2.2008}+2008/2009`

`=\sqrt{(2008+1)^2-2.2008+2008^2/2009^2}+2008/2009`

`=\sqrt{2009-2.2008/2009*2009+2008^2/2009^2}+2008/2009`

`=\sqrt{(2009-2008/2009)^2}+2008/2009`

`=|2009-2008/2009|+2008/2009`

`=2009-2008/2009+2008/2009`

`=2009` là 1 số tự nhiên

$a+b+c \ge \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$

$\Leftrightarrow 2a+2b+2c \ge 2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}$

$\Leftrightarrow a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+c-2\sqrt{ca}+a \ge 0$

$\Leftrightarrow (\sqrt{a}-\sqrt{b})^2+(\sqrt{c}-\sqrt{b})^2+(\sqrt{a}-\sqrt{c})^2 \ge 0$ luôn đúng với $a,b,c \ge 0$

Dấu "=" xảy ra khi a=b=c