giải phương trình: 1)2x+1/x-1=5(x-1)/x+1 2)x-2/2+x-1/x-2=2(x-11)/x^2-4 30/10/2021 Bởi Valerie giải phương trình: 1)2x+1/x-1=5(x-1)/x+1 2)x-2/2+x-1/x-2=2(x-11)/x^2-4
$\frac{2x+1}{(x-1)}$=$\frac{5(x-1)}{(x+1)}$ ⇔$\frac{(2x+1)(x+1)}{(x-1)(x+1)}$=$\frac{5(x-1)^{2}}{(x+1)(x-1)}$ ⇔$2x^{2}$ +x+2x+1=5($x^{2}$+2x+1) ⇔ $2x^{2}$ +x+2x+1=$5x^{2}$+10x+5 ⇔$3x^{2}$-7x-4=0 ⇔$3x^{2}$-3x-4x-4=0 ⇔3x(x-1)+4(x-1)=0 ⇔(x-1)(3x+4)=0 ⇔\(\left[ \begin{array}{l}x=1\\3x=-4\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=-\frac{4}{3}\end{array}\right.\) Bình luận
1, (2x+1)(x+1)=(5x-5)(x-1) (2x+1)(x+1)-(5x-5)(x-1)=0 2x^2+2x+x+1-5x^2+10x-5=0 -3x^2+13-4=0 3x^2-x-12x+4=0 x(3x-1)-4(3x-1)=0 (3x-1)(x-4)=0 Vậy x=1/3 hoặc x=4 Cho e ctlhn với Bình luận
$\frac{2x+1}{(x-1)}$=$\frac{5(x-1)}{(x+1)}$
⇔$\frac{(2x+1)(x+1)}{(x-1)(x+1)}$=$\frac{5(x-1)^{2}}{(x+1)(x-1)}$ ⇔$2x^{2}$ +x+2x+1=5($x^{2}$+2x+1)
⇔ $2x^{2}$ +x+2x+1=$5x^{2}$+10x+5 ⇔$3x^{2}$-7x-4=0
⇔$3x^{2}$-3x-4x-4=0 ⇔3x(x-1)+4(x-1)=0 ⇔(x-1)(3x+4)=0
⇔\(\left[ \begin{array}{l}x=1\\3x=-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-\frac{4}{3}\end{array}\right.\)
1, (2x+1)(x+1)=(5x-5)(x-1)
(2x+1)(x+1)-(5x-5)(x-1)=0
2x^2+2x+x+1-5x^2+10x-5=0
-3x^2+13-4=0
3x^2-x-12x+4=0
x(3x-1)-4(3x-1)=0
(3x-1)(x-4)=0
Vậy x=1/3 hoặc x=4
Cho e ctlhn với