D=|x-2010| + |x-2011| + |x-2012|
D=|x-2010| + |x-2011| + |2012-x|
=>D>=|x-2010+2012-x| + |x-2011|
=>D>=|2| + |x-2011|=2 + |x-2011|
Dấu = xảy ra <=> (x-2010)(2012-x)>=0<=>2010<=x<=2012(1)
                           x-2011=0 => x =2011(2)
Từ 1,2 => x=2011
Vậy Bmin=2 khi x=2011
 

Lời giải của mình ở đây nhé bạn!

http://olm.vn/hoi-dap/question/424173.html

moi hok lop 6

a: x>=9

=>x-9>=0

=>A=x-9+7=x-2

b: x<0

=>-3x>0

=>\(B=-3x-8x^2+x-2=-8x^2-2x-2\)

c: TH1: x>=1

=>x-1>=0

C=3x+|x-1|-2

=3x+x-1-2

=4x-3

TH2: x<1

=>x-1<0

C=3x+|x-1|-2

=3x-2-x+1

=2x-1

d: x<0

=>-x>0

\(D=\frac{x^2-\left|2x\right|+1}{\left(x+1\right)\cdot\left|-x\right|}\)

\(=\frac{x^2+2x+1}{\left(x+1\right)\cdot\left(-x\right)}\)

\(=\frac{\left(x+1\right)^2}{-x\left(x+1\right)}=\frac{x+1}{-x}\)

-2. (-1+x)+5. (-2+x)=-14+x

-2.(-1)+(-2).x + 5.(-2)+5.x=-14+x

2 -  2x -  10 +5x =-14+x

-2x+5x-x = -14+10 -2

 2x=-6

x=-6:2

x=-3

\(\frac{2009}{1}+\frac{2010}{2}+...+\frac{5016}{2008-2008}\)

\(=\frac{2009}{1}+\frac{2010}{2}+...+\frac{5016}{0}\)

Sau đó QĐM(bạn tự QĐ nha)

\(=\frac{0}{0}+\frac{0}{0}+...+\frac{5016}{0}\)

\(=\frac{5016}{0}=0\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right).x=0\)

Mà \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\ne0\)

\(\Rightarrow x=0\)

Ta có: \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=5\)

\(\Leftrightarrow x^2-2x+1-x^2+4=5\)

\(\Leftrightarrow-2x=0\)

hay x=0