\(P=\dfrac{x^2+x+1}{x^2+2x+1}\) ( x # -1)
\(P=\dfrac{\left(x+1\right)^2-x}{\left(x+1\right)^2}\)
\(P=1-\dfrac{x}{\left(x+1\right)^2}\)
\(P=1+\dfrac{1}{\left(x+1\right)^2}-\dfrac{1}{x+1}\)
\(P=\left[\dfrac{1}{\left(x+1\right)^2}-2.\dfrac{1}{x+1}.\dfrac{1}{2}+\dfrac{1}{4}\right]+1-\dfrac{1}{4}\)
\(P=\left(\dfrac{1}{x+1}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do : \(\left(\dfrac{1}{x+1}-\dfrac{1}{2}\right)^2\) ≥ 0 ∀x # -1
⇒ \(\left(\dfrac{1}{x+1}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) ≥ \(\dfrac{3}{4}\)
⇒ PMIN = \(\dfrac{3}{4}\) ⇔ x + 1 = 2 ⇔ x = 1
Mk làm cách khác nhé !!!
P = \(\dfrac{x^2+x+1}{x^2+2x+1}\)
P - 1 = \(\dfrac{x^2+x+1}{x^2+2x+1}\) - 1
P - 1 = \(\dfrac{-x}{x^2+2x+1}=\dfrac{-x}{x\left(x+2+\dfrac{1}{x}\right)}\)
P - 1 = \(\dfrac{-1}{x+\dfrac{1}{x}+2}\)
P - 1 = \(\dfrac{-1}{\left(\sqrt{x}-\sqrt{\dfrac{1}{x}}\right)^2+4}\) ≥ \(\dfrac{-1}{4}\)
⇒ P ≥ 1 - \(\dfrac{1}{4}=\dfrac{3}{4}\)
⇒ PMin = \(\dfrac{3}{4}\)
Dấu"=" xảy ra khi và chỉ khi : \(x=\dfrac{1}{x}\) ⇔ x = 1
Giải:
\(P=\dfrac{x^2+x+1}{x^2+2x+1}\)
\(\Leftrightarrow P=\dfrac{x^2+x+1}{\left(x+1\right)^2}\)
Vì \(\left(x+1\right)^2\ge0\)
Mà \(\left(x+1\right)^2\ne0\)
\(\Leftrightarrow\left(x+1\right)^2>0\)
\(\Leftrightarrow x>-1\)
Vậy ...
\(P=\dfrac{x^2+x+1}{x^2+2x+1}\\ =\dfrac{x^2+2x-x+1-1+1}{x^2+2x+1}\\ =\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\\ =1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\)
Đặt \(\dfrac{1}{x+1}=t\)
\(\Rightarrow P=t^2-t+1\\ =t^2-t+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(t^2-t+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(t-\dfrac{1}{2}\right)^2\ge0\forall x;t\)
\(\Rightarrow P=\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;t\)
Dấu "=" xảy ra khi:
\(\left(t-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow t-\dfrac{1}{2}=0\\ \Leftrightarrow t=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}\\ \Leftrightarrow x+1=2\\ \Leftrightarrow x=1\)
Vậy \(P_{Min}=\dfrac{3}{4}\) khi \(x=1\)
mk lại lm thêm cách khác nha :)
ta có : \(P=\dfrac{x^2+x+1}{x^2+2x+1}\) \(\Leftrightarrow P\left(x^2+2x+1\right)=x^2+x+1\)
\(\Leftrightarrow Px^2+2Px+P=x^2+x+1\Leftrightarrow x^2-Px^2+x-2Px+1-P=0\)
\(\Leftrightarrow\left(1-P\right)x^2+\left(1-2P\right)x+1-P=0\)
ta có : phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)
\(\Rightarrow\left(1-2P\right)^2-4\left(1-P\right)\left(1-P\right)\ge0\)
\(\Leftrightarrow4P^2-4P+1-4P^2+8P-4\ge0\Leftrightarrow4P-3\ge0\Leftrightarrow P\ge\dfrac{3}{4}\)
\(\Rightarrow\) giá trị nhỏ nhất của \(P\) là \(\dfrac{3}{4}\) khi \(x=\dfrac{-b}{2a}=\dfrac{-\left(1-2P\right)}{2\left(1-P\right)}=\dfrac{-\left(1-2.\dfrac{3}{4}\right)}{2\left(1-\dfrac{3}{4}\right)}=1\)
vậy \(P=\dfrac{x^2+x+1}{x^2+2x+1}\) đạt \(GTNN\) là \(\dfrac{3}{4}\) khi \(x=1\)
