Hoặc
b) 1x+2−1x−2=167
b) 1x+2−1x−2=167 x≠±2⇔x−2−x−2(x−2)(x+2)=167⇔−4x2−4=167⇔16x2−64=−28⇔16x2=36⇔x2=94⇒x=±32(tm)S=±32