Question

Tìm giá trị nhỏ nhất của biểu thức sau;

M=x^2+x+10

mai mik nộp rồi

Answer 1

\(M=x^2+x+10\)

\(=x^2+x+\frac{1}{4}+\frac{39}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\)

Vậy \(M_{min}=\frac{39}{4}\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Answer 2

\(M=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{39}{4}\)

\(M=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\)

\(\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\ge0\)​

​​\(\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\)

\(\Rightarrow M\ge\frac{39}{4}\)

Dấu "=" xảy ra: \(\left(x+\frac{1}{2}\right)^2=0\)

                            \(x+\frac{1}{2}=0\)

                            \(x=-\frac{1}{2}\)

Answer 3

\(M=x^2+x+10\)

\(M=x^2+x+\frac{1}{4}+\frac{39}{4}\)

\(M=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\)

Dấu '' = '' xảy ra

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

Vậy...........