Cho a,b,c tm a.b.c=1 Tính B= $\frac{1}{1+a+ab}$ + $\frac{1}{1+b+bc}$ + $\frac{1}{1+c+ca}$ 01/10/2021 Bởi Cora Cho a,b,c tm a.b.c=1 Tính B= $\frac{1}{1+a+ab}$ + $\frac{1}{1+b+bc}$ + $\frac{1}{1+c+ca}$
Ta có: ` B = \frac{1}{1 + a + ab} + \frac{1}{1 + b + bc} + \frac{1}{1 + c + ca} ` ` <=> B = \frac{1}{1 + a + ab} + \frac{abc}{abc + b + bc} + \frac{abc}{abc + c + ca} ` ` <=> B = \frac{1}{1 + a + ab} + \frac{ac}{ac + abc + c} + \frac{ab}{ab + 1 + a} ` ` <=> B = \frac{1}{1 + a + ab} + \frac{a}{a + ab + 1} + \frac{ab}{ab + 1 + a} ` ` <=> B = \frac{1 + a + ab}{1 + a + ab} ` ` <=> B = 1 ` Vậy ` B = 1 ` Bình luận
`B=1/(1+a+ab) +1/(1+b+bc)+1/(1+c+ca)` `⇔B=1/(1+a+ab)+(abc)/(abc+acb^2+bc)+(abc)/(abc+c+ca)` `⇔B=1/(1+a+ab)+a/(1+a+ab)+ab/(1+a+ab)` `⇔B=(1+a+ab)/(1+a+ab)` `⇔B=1` Bình luận
Ta có:
` B = \frac{1}{1 + a + ab} + \frac{1}{1 + b + bc} + \frac{1}{1 + c + ca} `
` <=> B = \frac{1}{1 + a + ab} + \frac{abc}{abc + b + bc} + \frac{abc}{abc + c + ca} `
` <=> B = \frac{1}{1 + a + ab} + \frac{ac}{ac + abc + c} + \frac{ab}{ab + 1 + a} `
` <=> B = \frac{1}{1 + a + ab} + \frac{a}{a + ab + 1} + \frac{ab}{ab + 1 + a} `
` <=> B = \frac{1 + a + ab}{1 + a + ab} `
` <=> B = 1 `
Vậy ` B = 1 `
`B=1/(1+a+ab) +1/(1+b+bc)+1/(1+c+ca)`
`⇔B=1/(1+a+ab)+(abc)/(abc+acb^2+bc)+(abc)/(abc+c+ca)`
`⇔B=1/(1+a+ab)+a/(1+a+ab)+ab/(1+a+ab)`
`⇔B=(1+a+ab)/(1+a+ab)`
`⇔B=1`