\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)

\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)

\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)

\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)

Tính giá trị biểu thức là " Nhân :hay " Chia " hay " Cộng" hay Trừ " vậy .

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2x+4}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

Khi \(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

Bài 2: 

a: \(A=\left(x+1\right)^3+5=20^3+5=8005\)

b: \(B=\left(x-1\right)^3+1=10^3+1=1001\)

Ta có: \(C=x^4-2x^3+3x^2-2x+2\)

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

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

\(=8x^2-8x+2+2\)

\(=8x^2-8x+4\)

Em tham khảo:

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

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

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

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

\(M=4.\left(4+2\right)+2\)( Vì \(x^2-x=4\))

\(M=24+2=26\)

Vậy M = 26 khi \(x^2-x=4\)