3x^2-2x+1 3x^4-8x^3-10x^2+8x-5 x^2-2x-16/3 3x^4-2x^3+x^2 -6x^3-12x^2+8x-5 -6x^3+4x^2-2x -16x^2+10x-5 -16x^2+32/3x-16/3 -2/3x+1/3

Vậy 

• (3x⁴-8x³-10x²+8x-5):(3x²-2x+1) = \(x^2-2x-\frac{16}{3}\)dư \(\frac{-2}{3}x+\frac{1}{3}\)

x^2-1 x^4-2x^3+2x-1 x^2-2x+1 x^4-x^2 -2x^3+x^2+2x-1 -2x^3+2x x^2-1 x^2-1 0

a) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x+4\right)\left(x-4\right)}{x\left(4-x\right)}\)

\(=\frac{\left(x+4\right)\left(x-4\right)}{-x\left(x-4\right)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\frac{x+1}{2}\)

c) \(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

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

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

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

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

d) \(\frac{x^3-x^2y+xy^2}{x^3+y^3}\)

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

lười :)))))))))))

did (ngu anh)

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

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

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

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.