\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{1c-1a+1a-1b+1b-1c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Sr nha

Kq mik nhầm

Ko phải -2b đâu mà = 0

\(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-x-2\right)=0\)

\(\Leftrightarrow-4\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\left(x-2\right)\left(x-2-x-2\right)=0\)

\(\left(x-2\right)\left(-4\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Qx=x^2-x-2x+2=x(x-1) - 2(x-1)

=(x-1)(x-2)  => Qx nhận x=1 và x=2 là nghiệm

theo định lý bowzu, để Px chia hết cho Qx thì P(1)=0 và P(2)=0

P(1)=1+a+b=0 =>a+b=-1  =>a=-1-b

P(2)=8+2a+b=0 => 2a+b=-8  => 2(-1-b) +b=-8

=>b-2-2b = -8

<=>-b=-6

<=>b=6

=>a = -1+6=5

vậy a=5, b=6

\(2x^2-x+5\)

\(=2\left(x^2-\frac{x}{2}+\frac{5}{2}\right)\)

\(2\left(x^2-2.x.\frac{1}{4}+\frac{1}{16}+\frac{39}{16}\right)\)

\(=2\left[\left(x-\frac{1}{4}\right)^2+\frac{39}{16}\right]\)

\(=2\left(x-\frac{1}{4}\right)^2+\frac{39}{8}\ge\frac{39}{8}\)

Dấu '' ='' xảy ra 

\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2\Leftrightarrow x=\frac{1}{4}\)

Vậy........

1²+5

=1+5

=6
 

\(B=2x^2+40x-1\)

\(=2\left(x^2+20x-\frac{1}{2}\right)\)

\(=2\left(x^2+20x+100-\frac{201}{2}\right)\)

\(=2\left[\left(x+10\right)^2-\frac{201}{2}\right]\)

\(=2\left[\left(x+10\right)^2\right]-201\ge-201\)

Vậy \(B_{min}=-201\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

\(B=2x^2+40x-1\)\(=2\left(x^2+2.10x+100\right)-201\)

\(=2\left(x+10\right)^2-201\)

Vì \(2\left(x+10\right)^2\ge0\forall x\)=>\(2\left(x+10\right)^2-201\ge-201\forall x\)

hay \(B\ge-201\forall x\)

\(MinB=-201\Leftrightarrow x+10=0\Leftrightarrow x=-10\)