Question

Cho biểu thức: \(P=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^{^2}-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^{^2}-3x}{2x^2-x^3}\)

a) Rút gọn A

b) Tìm các giá trị nguyên của x để \(P⋮4\)

c) Khi x>3. TÌm GTNN của P

Answer 1

a: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)

\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)

c: \(P=\dfrac{4x^2-12x+12x-36+36}{x-3}\)

\(=4x+12+\dfrac{36}{x-3}=4x-12+\dfrac{36}{x-3}+24\)

\(\Leftrightarrow P>=2\sqrt{4\left(x-3\right)\cdot\dfrac{36}{x-3}}+24=2\cdot\sqrt{144}+24=2\cdot12+24=48\)

Dấu = xảy ra khi 4(x-3)^2=36

=>(x-3)^2=9

=>x=6