\(A=\dfrac{x^3-27}{x-3}+5x\)
\(=\dfrac{\left(x-3\right)\left(x^2+3x+9\right)}{x-3}+5x\)
\(=x^2+3x+9+5x=x^2+8x+9\)
\(=x^2+8x+16-7\)
\(=\left(x+4\right)^2-7\)
Ta có: \(\left(x+4\right)^2\ge0\forall x\Rightarrow\left(x+4\right)^2-7\ge-7\forall x\)
Dấu "=" xảy ra khi x + 4 = 0 hay x = -4
Vậy AMAX = -7 khi x = -4.
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A = \(\dfrac{x^3-27}{x-3}+5x\)
= \(\dfrac{\left(x-3\right)\left(x^2-3x+9\right)}{x-3}+5x\)
= x2 + 3x + 9 + 5x
= x2 + 8x + 9
= x2 + 2x.4 + 8 + 9
= (x+4)2 + 9 \(\ge\) 9
Dấu "=" xảy ra khi x = -4
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