\(N=-x\left(x+1\right)-2y^2=-x^2-x-2y^2\)

\(=-x^2-x-\frac{1}{4}+\frac{1}{4}-2y^2\)

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

Vì \(-\left(x+\frac{1}{2}\right)^2\le0;-2y^2\le0\)

=> \(-\left(x+\frac{1}{2}\right)^2-2y^2+\frac{1}{4}\le\frac{1}{4}\)

=> \(N\le\frac{1}{4}\)

Dấu "=" xảy ra <=>  x = -1/2; y =0

Vậy max N = 1/4 tại x = -1/2 ; y = 0.

\(C=x^2+3x\)

\(=x^2+2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\ge0-\frac{9}{4};\forall x\)

Hay \(C\ge-\frac{9}{4};\forall x\)

Dấu"="xảy ra \(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=0\)

                     \(\Leftrightarrow x=\frac{-3}{2}\)

Vậy \(C_{min}=\frac{-9}{4}\)\(\Leftrightarrow x=\frac{-3}{2}\)

B=(x^2+x)(x^2+x-4)

Đặt a= x^2+x-2

=> B=(a+2)(a-2)=a^2-4

mà a^2>=0 => B>=-4

Dấu = xảy ra <=> a=0<=> x^2+x-2=0

<=> x^2-x+2x-2=0<=> x(x-1)+2(x-1)=0<=>(x-1)(x+2)=0 <=> x=1 hoặc -2

Vậy GTNN của B=-4 tại x=1 hoặc -2

Đặt A=x^4-x^3+3x^2-2x+2

=(x^4+3x^2+2)-(x^3+2x)

=(x^4+x^2+2x^2+2)-x(x^2+2)

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

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

Ta có x^2+2>=2>0;

x^2-x+1=(x^2-x+1/4)+3/4 =(x-1/2)^2+3/4>=3/4>0 

=> A>0  

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

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

\(\Rightarrow\frac{a^2}{b+c}+\frac{ab}{c+a}+\frac{ac}{a+b}+\frac{ab}{b+c}+\frac{b^2}{c+a}+\frac{bc}{a+b}\)

\(+\frac{ca}{b+c}+\frac{bc}{c+a}+\frac{c^2}{a+b}=a+b+c\)

\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{c\left(a+b\right)}{a+b}+\frac{a\left(b+c\right)}{b+c}\)

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

\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\left(a+b+c\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\left(đpcm\right)\)