Thực hiện phép tính: $\frac{x+3}{x+1}$- $\frac{2x-1}{x-1}$- $\frac{x-3}{x^{2}-1}$ 02/12/2021 Bởi Genesis Thực hiện phép tính: $\frac{x+3}{x+1}$- $\frac{2x-1}{x-1}$- $\frac{x-3}{x^{2}-1}$
Đáp án: `(x + 3)/(x + 1) – (2x – 1)/(x – 1) – (x – 3)/(x^2 – 1)` `= (x + 3)/(x + 1) – (2x – 1)/(x – 1) – (x – 3)/[(x – 1)(x + 1)]` `= [(x + 3)(x – 1)]/[(x – 1)(x + 1)] – [(2x – 1)(x + 1)]/[(x – 1)(x + 1)] – (x – 3)/[(x – 1)(x + 1)]` `= [(x + 3)(x – 1) – (2x – 1)(x + 1) – (x – 3)]/[(x – 1)(x + 1)]` `= (x^2 + 3x – x – 3 – 2x^2 + x – 2x + 1 – x + 3)/[(x – 1)(x + 1)]` `= (-x^2 + 1)/[(x – 1)(x + 1)]` `= [-(x – 1)(x + 1)]/[(x – 1)(x + 1)]` `= -1` Giải thích các bước giải: Bình luận
Đáp án:
`(x + 3)/(x + 1) – (2x – 1)/(x – 1) – (x – 3)/(x^2 – 1)`
`= (x + 3)/(x + 1) – (2x – 1)/(x – 1) – (x – 3)/[(x – 1)(x + 1)]`
`= [(x + 3)(x – 1)]/[(x – 1)(x + 1)] – [(2x – 1)(x + 1)]/[(x – 1)(x + 1)] – (x – 3)/[(x – 1)(x + 1)]`
`= [(x + 3)(x – 1) – (2x – 1)(x + 1) – (x – 3)]/[(x – 1)(x + 1)]`
`= (x^2 + 3x – x – 3 – 2x^2 + x – 2x + 1 – x + 3)/[(x – 1)(x + 1)]`
`= (-x^2 + 1)/[(x – 1)(x + 1)]`
`= [-(x – 1)(x + 1)]/[(x – 1)(x + 1)]`
`= -1`
Giải thích các bước giải: