Question

[Oriana.su]

ai giỏi toán giúp với

Answer 1

[Lê Thị Thục Hiền]

a) Đk:\(x\ne0;x\ne2\)

\(A=\left[\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^2\left(x-2\right)+4\left(x-2\right)}\right].\dfrac{x^2-x-2}{x^2}\)

\(=\left[\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right].\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x+1}{2x}\)

b)Tại \(x=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\) (tm đk) thay vào A ta được:

\(A=\dfrac{\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}}{2\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)\(=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{4}\)

Vậy...

Answer 2

[Oriana.su]

bạn ơi cái kết quả cuối cũng có thể nhân cái tử vào rồi rút gọn phân số luôn á