\(\Delta=b'^2-ac\)

\(\Delta=m^2-6m+12>0\forall m\in R\)

vậy pt luôn có 2 ngiệm \(x_1;x_2\)

theo vi-ét ta có

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=2m-4\\P=x_1x_2=\dfrac{c}{a}=2m-8\end{matrix}\right.\)

ta có \(P=\left(x_1-x_2\right)^2=S^2-4P=4m^2-24m+48\)

\(P=4\left(m-3\right)^2+12\ge12\forall m\in R\)

dấu = xảy ra khi m=3

Vậy \(P_{min}=12\) khi m=3

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

\(2M=x^2-2xy+y^2+x^2-2x+1+y^2+2y+1\)

\(M=\dfrac{\left(x-y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2}{2}\ge0\)

\(=-10x^3\left(\dfrac{2}{5}x^2y+\dfrac{3}{10}xy^2\right)+3x^4y^3\)

\(=-4x^5y-3x^4y^2+3x^4y^3\)

a) 8x(x-2013)-(x-2013)=0

\(\Leftrightarrow\left(x-2013\right)\left(8x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=\dfrac{1}{8}\end{matrix}\right.\)

c/m \(\left(a+4b\right)^3\ge81ab^2\)

theo cosi ta co

\(\left(a+b+3b\right)^3\ge\left(3\sqrt[3]{a.3b^2}\right)^3=81ab^2\)

biết bỏ ngoặc mà sao ko làm

nếu ko làm như vậy thì chẳng còn cách làm

B=\(\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)....\left(1-\dfrac{7}{7}\right)...\left(1-\dfrac{2011}{7}\right)\)

B=\(\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)...0...\left(1-\dfrac{2011}{7}\right)=0\)

12 cm

tich nha cảm ơn thật đấy