(x+2)²-x²+2x-1

=(x+2)²-(x²-2x+1)

=(x+2)² - ( x-1)² 

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

=3(2x+1)

\(\left(x+2\right)^2-x^2+2x-1\)

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

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

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

\(=[\left(x+2\right)-\left(x-1\right)].[\left(x+2\right)+\left(x-1\right)]\)

\(=\left(x+2+x-1\right).\left(x+2-x+1\right)\)

\(=3.\left(2x+1\right)\)

\(\left(x+3\right)^2+\left(x-3\right)^2-2.\left(x+3\right).\left(x-3\right)\)

\(=\left(x^2+6x+9\right)+\left(x^2-6x+9\right)-2.\left(x^2-9\right)\)

\(=x^2+6x+9+x^2+x^2-6x+9-2x^2+18\)

\(=36\)

\(x+2y=3\)

\(\Rightarrow x=3-2y\)

\(S=x^2+y^2\)

\(=\left(3-2y\right)^2+y^2\)

\(=9-12y+4y^2+y^2\)

\(=5y^2-12y+9\)

\(=5.\left(y^2-\frac{12}{5}y+\frac{9}{5}\right)\)

\(=5.\left(y^2-2y.\frac{6}{5}+\frac{36}{25}+\frac{9}{25}\right)\)

\(=5.\left(\frac{-6}{5}\right)^2+\frac{9}{5}\ge\frac{9}{5}\forall y\)

Dấu '' = '' xảy ra khi: \(\left(y-\frac{6}{5}\right)^2=0\Rightarrow y=\frac{6}{5}\)

Do vậy: \(x=3-2.\frac{6}{5}=\frac{3}{5}\)

Vậy giá trị nhỏ nhất của \(S=\frac{9}{5}\) khi \(x=\frac{3}{5}\) và \(y=\frac{6}{5}\)

\(x+2y=3=>x=3-2y\)thay vào S ta được:\(S=\left(3-2y\right)^2+y^2=5y^2-12y+9=y^2-\frac{12}{5}y+\frac{9}{5}=\left(y-\frac{6}{5}\right)^2+\frac{9}{25}\ge\frac{9}{25}\)

Dấu "=" xảy ra khi \(y-\frac{6}{5}=0< =>y=\frac{6}{5}\)

Vậy Min \(S=\frac{9}{25}\)khi \(y=\frac{6}{5}\)

mình nghĩ là làm vậy

đường trung bình  \(MN=\frac{1}{2}.\left(AB+DC\right)\)thì phải, mình ko nhớ lắm