Toán Lớp 8: tìm gtnn a, (x-2)^2+(x-1)^2+5
b,x^2-x-1
c, a^4-2a^3+3a^2-4a+5
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Toán Lớp 8: tìm gtnn a, (x-2)^2+(x-1)^2+5
b,x^2-x-1
c, a^4-2a^3+3a^2-4a+5
Comments ( 2 )
⇒ (x-2)² + (x-1)² + 5 ≥ 5
Dấu “=” xảy ra khi (x-2)², (x-1)² = 0
⇒ \(\left[ \begin{array}{x}x-2 = 0\\x-1=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{x}x=2\\x=1\end{array} \right.\)
b) x² – x – 1 = x² – x + $\frac{1}{4}$ – $\frac{5}{4}$
= (x – $\frac{1}{2}$ )² – $\frac{5}{4}$
║ (x – $\frac{1}{2}$ )² ≥ 0
⇒ (x – $\frac{1}{2}$ )² – $\frac{5}{4}$ ≥ – $\frac{5}{4}$
Dấu “=” xảy ra khi (x – $\frac{1}{2}$ )² = 0
⇒ x = $\frac{1}{2}$
c) $a^{4}$ – 2a³ + 2a² – 4a + 5
= ($x^{4}$ – 2a³ + a²) + (2a² – 4a +2) + 3
= (a²-a)² + ( $\sqrt[]{2}$ a – $\sqrt[]{2}$ )² + 3
║ (a²-a)² ≥ 0, ( $\sqrt[]{2}$ a – $\sqrt[]{2}$ )² ≥ 0
⇒ (a²-a)² + ( $\sqrt[]{2}$ a – $\sqrt[]{2}$ )² + 3 ≥ 3
Dấu “=” xảy ra khi
\(\left[ \begin{array}{1}a²-a = 0\\ \sqrt[]{2} a – \sqrt[]{2} =0\end{array} \right.\)
⇒ a = 1
____Cho mình xin ctlhn nhé, chúc bạn học tốt !!!_________