Tìm GTNN: A= (x+1)(x+2)(x+3)(x+4)-32 B= x^2-2xy+y^2+3x-3y+1 C=4x^2+1/x^2-20 (x>0)

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1. a) $A=(x+1)(x+2)(x+3)(x+4)-32$

   ⇔ $A=(x+1)(x+4)(x+2)(x+3)-32$

   ⇔ $A=(x^2+x+4x+4)(x^2+2x+3x+6)-32$

   ⇔ $A=(x^2+5x+4)(x^2+5x+4+2)-32$

   ⇔ $A=(x^2+5x+4)^2+2(x^2+5x+4)+1-33$

   ⇔ $A=(x^2+5x+4+1)^2-33$

   ⇔ $A=(x^2+5x+5)^2-33$

   Vì $(x^2+5x+5)^2 \geq 0$

   nên $(x^2-5x+5)^2-33 \geq -33$

   Dấu “=” xảy ra khi $x^2-5x+5=0$

   ⇔ $4x^2-20x+20=0$

   ⇔ $4x^2-20x+25-5=0$

   ⇔ $(2x-5)^2-5=0$

   ⇔ $(2x-5+\sqrt{5})(2x-5-\sqrt{5})=0$

   ⇔ \(\left[ \begin{array}{l}x=\dfrac{5-\sqrt{5}}{2}\\x=\dfrac{5+\sqrt{5}}{2}\end{array} \right.\)

    

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