Trang chủ » Hỏi đáp » Môn Toán 2x/x-2 + 1/x+2 – x^2-8/x^2-4 2x/x-5 + 10/5-x 10/08/2024 2x/x-2 + 1/x+2 – x^2-8/x^2-4 2x/x-5 + 10/5-x
(2x)/(x-2)+1/(x+2)-(x^2-8)/(x^2-4)(ĐK:x\ne2; x\ne-2) =(2x.(x+2)+(x-2)-(x^2-8))/((x-2).(x+2)) =(2x^2+4x+x-2-x^2+8)/((x-2).(x+2)) =(x^2+5x+6)/((x-2).(x+2)) =(x^2+2x+3x+6)/((x-2).(x+2)) =(x.(x+2)+3.(x+2))/((x-2).(x+2)) =((x+2).(x+3))/((x-2).(x+2)) =(x+3)/(x-2) $\\$ (2x)/(x-5)+10/(5-x)(ĐK:x\ne5) =(2x)/(x-5)-10/(x-5) =(2x-10)/(x-5) =(2.(x-5))/(x-5) =2(tm) Trả lời
1 bình luận về “2x/x-2 + 1/x+2 – x^2-8/x^2-4 2x/x-5 + 10/5-x”