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)
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)
By Natalia
By Natalia
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)
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 $x^2-5x+5=0$
⇔ $4x^2-20x+20=0$
⇔ $4x^2-20x+25-5=0$
⇔ $(2x-5)^2-5=0$
⇔ $(2x-5+\sqrt{5})(2x-5-\sqrt{5})=0$
⇔ \(\left[ \begin{array}{l}x=\dfrac{5-\sqrt{5}}{2}\\x=\dfrac{5+\sqrt{5}}{2}\end{array} \right.\)