Toán `((x+1)/(2x-2) + 3/(x^2 -1) – (x+3)/(2x+2)) . (4x^2 – 4)/5` 10/07/2021 By Isabelle `((x+1)/(2x-2) + 3/(x^2 -1) – (x+3)/(2x+2)) . (4x^2 – 4)/5`
Đáp án: Giải thích các bước giải: `((x+1)/(2x-2) + 3/(x^2 -1) – (x+3)/(2x+2)) . (4x^2 – 4)/5` ĐK: `x \ne +-1` `=[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x-1)(x+1)}].\frac{4(x^2-1)}{5}` `=[\frac{x^2+2x+1+6-x^2-2x+3}{2(x-1)(x+1)}].\frac{4(x-1)(x+1)}{5}` `=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}` `=4` Trả lời
Đáp án + Giải thích các bước giải: `((x+1)/(2x-2)+(3)/(x^{2}-1)-(x+3)/(2x+2)).(4x^{2}-4)/(5)` `=((x+1)/(2(x-1))+(3)/((x-1)(x+1))-(x+3)/(2(x+1))).(4(x^{2}-1))/(5)` `=((x+1)^{2}+6-(x+3)(x-1))/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)` `=(x^{2}+2x+1+6-(x^{2}+3x-x-3))/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)` `=(x^{2}+2x+7-(x^{2}+2x-3))/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)` `=(x^{2}+2x+7-x^{2}-2x+3)/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)` `=(10)/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)` `=4` Trả lời
Đáp án:
Giải thích các bước giải:
`((x+1)/(2x-2) + 3/(x^2 -1) – (x+3)/(2x+2)) . (4x^2 – 4)/5`
ĐK: `x \ne +-1`
`=[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x-1)(x+1)}].\frac{4(x^2-1)}{5}`
`=[\frac{x^2+2x+1+6-x^2-2x+3}{2(x-1)(x+1)}].\frac{4(x-1)(x+1)}{5}`
`=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}`
`=4`
Đáp án + Giải thích các bước giải:
`((x+1)/(2x-2)+(3)/(x^{2}-1)-(x+3)/(2x+2)).(4x^{2}-4)/(5)`
`=((x+1)/(2(x-1))+(3)/((x-1)(x+1))-(x+3)/(2(x+1))).(4(x^{2}-1))/(5)`
`=((x+1)^{2}+6-(x+3)(x-1))/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)`
`=(x^{2}+2x+1+6-(x^{2}+3x-x-3))/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)`
`=(x^{2}+2x+7-(x^{2}+2x-3))/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)`
`=(x^{2}+2x+7-x^{2}-2x+3)/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)`
`=(10)/(2(x-1)(x+1)).(4(x-1)(x+1))/(5)`
`=4`