a) \(A=9x^2-30x+30\)
\(A=\left(3x\right)^2-2\cdot3x\cdot5+5^2+5\)
\(A=\left(3x-5\right)^2+5\ge5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{3}\)
b) \(B=16x^2-24x-3\)
\(B=\left(4x\right)^2-2\cdot4x\cdot3+3^2-13\)
\(B=\left(4x-3\right)^2-13\ge-13\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)
c) \(C=4x^2+40x-2\)
\(C=\left(2x\right)^2+2\cdot2x\cdot10+10^2-102\)
\(C=\left(2x+10\right)^2-102\ge-102\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-5\)
d) \(D=36x^2-24x+7\)
\(D=\left(6x\right)^2-2\cdot6x\cdot2+2^2+3\)
\(D=\left(6x-2\right)^2+3\ge3\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{3}\)
a. A = \(9x^2-30x+30\)
= \(\left[\left(3x\right)^2+2.3x.5+5^2\right]+5\)
= \(\left(3x+5\right)^2+5\)
Ta có: \(\left(3x+5\right)^2\ge0\forall x\Rightarrow\left(3x+5\right)^2+5\ge5\forall x\)
Dấu "=" xảy ra khi: \(\left(3x+5\right)^2=0\)
\(\Leftrightarrow3x+5=0\Leftrightarrow x=-\frac{5}{3}\)
Vậy Min A = 5 \(\Leftrightarrow x=-\frac{5}{3}\)
b. B = \(16x^2-24x-3\)
= \(\left[\left(4x\right)^2-2.4x.5+5^2\right]-28\)
= \(\left(4x-5\right)^2-28\)
Ta có: \(\left(4x-5\right)^2\ge0\forall x\)
\(\Rightarrow\left(4x-5\right)^2-28\ge-28\forall x\)
Dấu "=" xảy ra khi: \(\left(4x-5\right)^2=0\)
\(\Leftrightarrow4x-5=0\Leftrightarrow x=\frac{5}{4}\)
Vậy Min B = -28 \(\Leftrightarrow x=\frac{5}{4}\)
b. B = 16x2−24x−316x2−24x−3
= [(4x)2−2.4x.5+52]−28[(4x)2−2.4x.5+52]−28
= (4x−5)2−28(4x−5)2−28
Ta có: (4x−5)2≥0∀x(4x−5)2≥0∀x
⇒(4x−5)2−28≥−28∀x⇒(4x−5)2−28≥−28∀x
Dấu "=" xảy ra khi: (4x−5)2=0(4x−5)2=0
⇔4x−5=0⇔x=54⇔4x−5=0⇔x=54
Vậy Min B = -28 ⇔x=54
