Ta có \(\left(x+1\right)\left(x^2+x+1\right)=0\)=>\(\orbr{\begin{cases}x+1=0\\x^2+x+1=0\end{cases}}\)Mà x^2+x+1>=0 với mọi x =>x=-1

\(x^3+2x^2+2x+1=0\)

\(\Leftrightarrow\left(x^3+1\right)+\left(2x^2+2x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)=0\)

Ta có: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\Rightarrow x+1=0\Leftrightarrow x=-1\)

Có P = x² + 5y² + 4xy + 6x + 16y + 32

         = [(x² + 4xy + 4y²) + 6x + 12y + 9] + (y² + 4y + 2²) + 19

         = [(x + 2y)² + 2(x + 2y).3 + 3² ] + (y + 2)² + 19

         = (x + 2y + 3)² + (y + 2)² + 19

Thấy (x + 2y + 3)² ≥ 0 với mọi x; y

         (y + 2)² ≥ 0 với mọi y

=> (x + 2y + 3)² + (y + 2)² ≥ 0 với mọi x; y

=> (x + 2y + 3)² + (y + 2)² + 19 ≥ 19 với mọi x; y

=> P ≥ 19 với mọi x; y

Dấu "=" xảy ra khi x + 2y + 3 = 0 và y + 2 = 0

Bn tự giải tiếp nha, mk ko biết có nhầm chỗ nào ko nhưng cách lm như vậy đó

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) \(\frac{x^3-x^2y+xy^2}{x^3+y^3}\)

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