a. Ta có:
\(A=2x^2-8x+10\\ =2\left(x^2-4x+5\right)=2\left[\left(x^2-2.x.2+4\right)+1\right]\\ =2\left[\left(x-2\right)^2+1\right]\\ =2\left(x-2\right)^2+2\)
Vì \(\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2\ge0\forall x\\ \Leftrightarrow2\left(x-2\right)^2+1\ge1\forall x\)
Dấu = xảy ra khi: \(2\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(MinA=1\Leftrightarrow x=2\)
GTLN
\(A=2x^2-8x-10\)
\(A=2\left(x^2-4x-5\right)\)
\(A=2\left(x^2-2.x.2+2^2-2^2-5\right)\)
\(A=2\left[\left(x^2-4x+2^2\right)-4-5\right]\)
\(A=2\left(x-2\right)^2-9\)
suy ra \(\left(x-2\right)^2\ge-9\)
=> Min A = -9 khi \(\left(x-2\right)^2=0\Leftrightarrow x-2=0hayx=2\)
Vậy Min A = (-9) khi x =2
B= 3.(x2 + 3x)
= 3. ( x2 + 2. \(\frac{3}{2}\).x + \(\frac{9}{4}\)) -\(\frac{9}{4}\)
= 3. (x+\(\frac{3}{2}\))2-\(\frac{9}{4}\)
Ta có: 3.(x+\(\frac{3}{2}\))2 ≥ 0 vs mọi x
<=> 3.(x+\(\frac{3}{2}\))2-\(\frac{9}{4}\) ≤ -\(\frac{9}{4}\)
Dấu bằng xảy ra <=> x+\(\frac{3}{2}\)= 0
=> x = -\(\frac{3}{2}\)
Vậy vs x = -\(\frac{3}{2}\) thì max của B =-\(\frac{9}{4}\)
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