Question

Tìm GTNN của :

a)N=( x-3)^2+6

b) P= 2.(x-5/4)^2+39/8

Answer 1

a, ta có :(x-3)\(^2\)\(\ge\)0 nên N \(\ge\)6

Min N =6 khi (x-3)\(^2\)=0\(\Rightarrow\)x-3=0\(\Rightarrow\)x=3

Vậy Min N=6 với x=3

b, vì (x-\(^{\frac{5}{4}}\))\(^2\)\(\ge\)0 nên 2(x-\(\frac{5}{4}\))\(\ge\)0\(\Rightarrow\) P \(\ge\frac{39}{8}\)

Min P=\(\frac{39}{8}\) với 2(x-\(\frac{5}{4}\))=0\(\Rightarrow\)x-\(\frac{5}{4}\)=0\(\Rightarrow x=\frac{5}{4}\)

vậy Min P=\(\frac{39}{8}\) với x =\(\frac{5}{4}\)

 

Answer 2

a)Ta thấy: \(\left(x-3\right)^2\ge0\)

\(\Rightarrow\left(x-3\right)^2+6\ge6\)

\(\Rightarrow N\ge6\)

Dấu "=" khi \(x=3\)

Vậy \(Min_N=6\) khi \(x=3\)

b)Ta thấy: \(\left(x-\frac{5}{4}\right)^2\ge0\)

\(\Rightarrow2\left(x-\frac{5}{4}\right)^2\ge0\)

\(\Rightarrow2\left(x-\frac{5}{4}\right)^2+\frac{39}{8}\ge\frac{39}{8}\)

Dấu "=" khi \(x=\frac{5}{4}\)

Vậy \(Min_P=\frac{39}{8}\) khi \(x=\frac{5}{4}\)