Question

Văn Tường ơi giúp tớ với!!

Answer 1

a)\(B=\dfrac{a^2}{a-1}+\left(\dfrac{a}{a^2-1}+\dfrac{1}{a-a^3}\right):\dfrac{1-a}{a+a^3}\)

\(B=\dfrac{a^2}{a-1}+\dfrac{a^2-1}{a\left(a^2-1\right)}.\dfrac{a+a^3}{1-a}\)

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

\(B=\dfrac{a^2}{a-1}+\dfrac{1+a^2}{1-a}\)

\(B=\dfrac{a^2}{a-1}+\dfrac{-1-a^2}{a-1}\)

\(B=\dfrac{-1}{a-1}\)

b)\(2a^2=a\Leftrightarrow2a^2-a=0\Leftrightarrow a\left(2a-1\right)=0\)<=>a=0 hoặc 2a-1=0

<=>a=0 hoặc a=1/2

TH1: a=0 => \(B=\dfrac{-1}{a-1}=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

TH2: a=1/2 => \(B=\dfrac{-1}{a-1}=\dfrac{-1}{\dfrac{1}{2}-1}=\dfrac{-1}{-\dfrac{1}{2}}=2\)

c)\(B=\dfrac{-1}{a-1}< 1\Leftrightarrow-1< a-1\Leftrightarrow a>0\)

Vậy B<1 khi a>0