A= \(\dfrac{x^2-4x+1}{x^2}\)
ĐKXĐ x≠0
A= \(\dfrac{x^2}{x^2}-\dfrac{4x}{x^2}+\dfrac{1}{x^2}\)
=\(1-\dfrac{4}{x}+\dfrac{1}{x^2}\)
đặt \(\dfrac{1}{x}=y\) ta có
1-4y+y2
= y2-4y+1
=(y2-4y+4)-3
= (y-2)2 -3
do (y-2)2 ≥ 0 ∀x
=> (y-2)2 -3 ≥ -3
=> A ≥ -3
=> Amin =-3dấu '=' xảy ra khi
y-2=0
=> y=2
=> \(\dfrac{1}{x}=2\)
=> x=\(\dfrac{1}{2}\)
vậy GTNN A =-3 khi x=\(\dfrac{1}{2}\)
b) ĐKXĐ x ≠\(\dfrac{1}{2}\)
B = \(\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\)
=\(\dfrac{4x^2-6x+1-1+1}{\left(2x-1\right)^2}\)
= \(\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\)
=\(\dfrac{\left(2x-1\right)^2-\left(2x-1\right)-1}{\left(2x-1\right)^2}\)
= \(\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\)
= \(1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\)
đặt \(-\dfrac{1}{2x-1}=y\) ta có
1+y+y2
= \(y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\)
=\(\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
do \(\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\)
=> \(\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
=> B ≥\(\dfrac{3}{4}\)
GTNN B =\(\dfrac{3}{4}\)dấu '=' xảy ra khi
y=-\(\dfrac{1}{2}\)
⇔\(-\dfrac{1}{2x-1}=-\dfrac{1}{2}\)
⇔2x-1=2
⇔2x=3
⇔x=\(\dfrac{3}{2}\) (tm)
vậy GTNN B=\(\dfrac{3}{4}\) khi x= \(\dfrac{3}{2}\)
