Câu hỏi của Thương Thương

Question

Giải các phương trình sau:

1) $\frac{x+3}{x+1}+\frac{x-2}{x}=2$

2) $\frac{2x-3}{x-2}=\frac{2x+1}{x}$

Ngọc Huyền · Accepted Answer

1. $\frac{x+3}{x+1}+\frac{x-2}{x}=2$         DKXD : $x\ne-1;x\ne0$

↔ $\frac{x\left(x+3\right)}{x\left(x+1\right)}+\frac{\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}$

→ $x\left(x+3\right)+\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)$

↔ $x^2+3x+x^2+x-2x-2=2x^2+2x$

↔ $x^2+3x+x^2+x-2x-2x^2-2x=2$

↔ $0x=2$

Ptr vo nghiem

2.$\frac{2x-3}{x-2}=\frac{2x+1}{x}$ DKXD : $x\ne0;x\ne2$

↔ $\frac{x\left(2x-3\right)}{x\left(x-2\right)}=\frac{\left(2x+1\right)\left(x-2\right)}{x\left(x-2\right)}$

→ $x\left(2x-3\right)=\left(2x+1\right)\left(x-2\right)$

↔ $2x^2-3x=2x^2-4x+x-2$

↔ $2x^2-3x-2x^2+4x-x=-2$

↔ $0x=-2$

Ptr vo nhiemCâu trả lời: 1. $\frac{x+3}{x+1}+\frac{x-2}{x}=2$         DKXD : $x\ne-1;x\ne0$