Question

Cho biểu thức

M= \(18+4x-8y+6xy+5x^2+10y^2\)

Chứng minh M > 0 với mọi giá trị của x,y

Answer 1

\(M=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)

\(M=\left(x+3y\right)^2+\left(2x+1\right)^2+\left(y-4\right)^2+1>0;\forall x;y\)

Answer 2

Ta có: \(M=18+4x-8y+6xy+5x^2+10y^2\)

\(=4x^2+4x+1+x^2+6xy+9y^2+y^2-8y+16+1\)

\(=\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\)

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\left(x+3y\right)^2\ge0\forall x,y\)

\(\left(y-4\right)^2\ge0\forall y\)

Do đó: \(\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\ge1>0\forall x,y\)

hay \(M>0\forall x,y\)