Question

Answer 1

1: \(2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

=>(2x+3)(x-1)=0

=>x=-3/2 hoặc x=1

Thay x=-3/2 vào B, ta được:

\(B=\left(\dfrac{9}{4}+\dfrac{3}{2}\right):\left(2\cdot\dfrac{-3}{2}+1\right)=\dfrac{15}{4}:\left(-2\right)=-\dfrac{15}{8}\)

Thay x=1 vào B, ta được:

\(B=\dfrac{1^2-1}{2\cdot1+1}=0\)

2: \(A=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

3: \(M=A\cdot B=\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}=\dfrac{x}{x+1}\)

Để M là số nguyên thì \(x+1-1⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1\right\}\)

hay \(x\in\left\{0;-2\right\}\)