P = 1 + x 1 2x
( ———- ) : ( —— – ———————- )
x^2 + 1 x – 1 x^3 + x – x^2 – 1
cần gấp trong 5p anh azzken giup em
P = 1 + x 1 2x ( ———- ) : ( —— – ———————- ) x^2 + 1
By Valerie
Đáp án:
a, Ta có :
`P = (x + 1)/(x^2 + 1) : (1/(x – 1) – (2x)/(x^3 + x – x^2 – 1) )`
`= (x + 1)/(x^2 + 1) : (1/(x – 1) – (2x)/[(x^2 + 1)(x – 1)] )`
`= [(x + 1)(x – 1)]/[(x^2 + 1)(x – 1)] : ( (x^2 + 1)/[(x – 1)(x^2 + 1)] – (2x)/[(x^2 + 1)(x – 1)] )`
` = [(x + 1)(x – 1)]/[(x^2 + 1)(x – 1)] : (x^2 – 2x + 1)/[(x^2 + 1)(x – 1)]`
`= [(x + 1)(x – 1)]/[(x^2 + 1)(x – 1)] . [(x^2 + 1)(x – 1)]/(x^2 – 2x + 1)`
`= [(x + 1)(x – 1)]/(x^2 – 2x + 1)`
`= [(x + 1)(x – 1)]/(x – 1)^2`
`= (x + 1)/(x – 1)`
b, Thay `x = -1/2` vào P ta được :
`P = (-1/2 + 1)/(-1/2 – 1) = (1/2)/(-3/2) = 1/(-3)`
Giải thích các bước giải:
`x^3 + x – x^2 – 1`
`= (x^3 + x) – (x^2 + 1)`
`= x(x^2 + 1) – (x^2 + 1)`
`= (x^2 + 1)(x – 1)`