Question

Tìm GTNN:

a. B= ( x-1)² + (x+3)² + ( x+5)²

b. C= ( x+5)⁴ + ( x+1)⁴

Answer 1

Lời giải:

a)

\(B=(x-1)^2+(x+3)^2+(x+5)^2=(x^2-2x+1)+(x^2+6x+9)+(x^2+10x+25)\)

\(=3x^2+14x+35=3(x^2+\frac{14}{3}x+\frac{7^2}{3^2})+\frac{56}{3}\)

\(=3(x+\frac{7}{3})^2+\frac{56}{3}\geq \frac{56}{3}\)

Dấu "=" xảy ra khi \((x+\frac{7}{3})^2=0\Leftrightarrow x=\frac{-7}{3}\)

Vậy \(B_{\min}=\frac{56}{3}\Leftrightarrow x=\frac{-7}{3}\)

b) Đặt $x+3=a$. Khi đó:

\(C=(x+5)^4+(x+1)^4=(a+2)^4+(a-2)^4\)

\(=2a^4+28a^2+32\geq 2.0+28.0+32=32\)

Dấu "=" xảy ra khi \(a^4=a^2=0\Leftrightarrow a=0\Leftrightarrow x=-3\)

Vậy \(C_{\min}=32\Leftrightarrow x=-3\)