Question

Cho biểu thức P=\(\dfrac{a^2+a}{a^2-2a+1}\):(\(\dfrac{a+1}{a}+\dfrac{1}{a-1}\)\(+\)\(\dfrac{2-a^2}{a^2-a}\))

a, Rút gọn P

b, Tìm a để P=\(\dfrac{-1}{2}\)

c, Tìm GTNN của P khi a>1

Answer 1

 

ĐKXĐ: a<>1; a<>0; a<>-1

a: \(P=\dfrac{a\left(a+1\right)}{\left(a-1\right)^2}:\dfrac{a^2-1+a+2-a^2}{a\left(a-1\right)}\)

\(=\dfrac{a\left(a+1\right)}{\left(a-1\right)^2}\cdot\dfrac{a\left(a-1\right)}{a+1}=\dfrac{a^2}{a-1}\)

b: Khi P=-1/2 thì a^2/(a-1)=-1/2

=>2a^2=-a+1

=>2a^2+a-1=0

=>2a^2+2a-a-1=0

=>(a+1)(2a-1)=0

=>a=1/2(nhận) hoặc a=-1(loại)

c: \(P=\dfrac{a^2-1+1}{a-1}=a+1+\dfrac{1}{a-1}=a-1+\dfrac{1}{a-1}+2\)

=>\(P>=2\cdot\sqrt{\left(a-1\right)\cdot\dfrac{1}{a-1}}+2=4\)

Dấu = xảy ra khi a-1=1

=>a=2