Question

Cho biểu thức A = \(\frac{1}{x-2}\) + \(\frac{1}{x+2}\) +\(\frac{x^2+1}{x^2-4}\)

a, Rút gọn

 

 

Answer 1

Đào Thị Phương Duyên mk trả lời câu kia của pn rùi ak

Answer 2

nhẩm cx ra (x+1)² / x² -4

Answer 3

Giải

x-2

x+2

x² - 4 = (x-2).(x+2)

=> MTC : (x-2).(x+2)

NTP : NTP1:x+2

NTP2:x-2

NTP3:1

Ta có :

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

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

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

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

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

 

(mik lm thế, nếu có sai thì mong bn thông cảm nha ^^)