2+$\frac{2x^2-8}{2x^2+8}$+$\frac{2x^2+7+23}{2x^2+7-4}$=$\frac{2x+5}{2x-1}$ Giải PT: 05/11/2021 Bởi Genesis 2+$\frac{2x^2-8}{2x^2+8}$+$\frac{2x^2+7+23}{2x^2+7-4}$=$\frac{2x+5}{2x-1}$ Giải PT:
ta cos” ⇔2+2x(x−4)/2x(x+4)+(2x^2+7x+23)/(2x−1)(x+4)=2x+5/2x−1 ⇔2+x−4/x+4+(2x^2+7x+23)/(2x−1)(x+4)−2x+5/2x−1=0 ⇔{2(x+4)(2x−1)+(x−4)(2x−1)+2x^2+7x+23−(2x+5)(x+4)}/(x+4)(2x−1)=0 ⇔2(x+4)(2x−1)+(x−4)(2x−1)+2x^2+7x+23−(2x+5)(x+4)=0 ⇔4x^2+14x−8+2x^2−9x+4+2×62+7x+23−2x^2−13x−20=0 ⇔6x^2+7x−1=0 ⇔6(x^2+2.7/12.x+49/144)−193/144=0 ⇔(x+7/12)^2=193/144/6=193/864 banj lamf nots nhes g/o/g/o Bình luận
ta cos”
⇔2+2x(x−4)/2x(x+4)+(2x^2+7x+23)/(2x−1)(x+4)=2x+5/2x−1
⇔2+x−4/x+4+(2x^2+7x+23)/(2x−1)(x+4)−2x+5/2x−1=0
⇔{2(x+4)(2x−1)+(x−4)(2x−1)+2x^2+7x+23−(2x+5)(x+4)}/(x+4)(2x−1)=0
⇔2(x+4)(2x−1)+(x−4)(2x−1)+2x^2+7x+23−(2x+5)(x+4)=0
⇔4x^2+14x−8+2x^2−9x+4+2×62+7x+23−2x^2−13x−20=0
⇔6x^2+7x−1=0
⇔6(x^2+2.7/12.x+49/144)−193/144=0
⇔(x+7/12)^2=193/144/6=193/864
banj lamf nots nhes
g/o/g/o