\(a.P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)=\dfrac{x^2+x}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}=\dfrac{x\left(x+1\right)}{x-1}.\dfrac{x}{x+1}=\dfrac{x^2}{x-1}\) ( x # 1 ; x # 0 )
\(b.P=\dfrac{1}{2}\) ⇔ \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\) ⇔ 2x2 = x - 1 ⇔ 2x2 - x + 1 = 0
Ta thấy : \(2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+1-\dfrac{1}{8}=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}>0\)
Vậy , phương trình trên vô nghiệm .
\(c.P-2=\dfrac{x^2}{x-1}-2=\dfrac{x^2-2x+2}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(x-1+\dfrac{1}{x-1}\) ≥ \(2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}=2\)
⇔ \(x-1+\dfrac{1}{x-1}+2\) ≥ \(4\)
⇔ \(P_{Min}=4\) ⇔ x = 2
