Bài `1 :`

`b)` Ta có `: a/( ab + a + 2017 ) = a/( ab + a + abc ) = 1/( b + 1 + bc )` 

                  `( 2017 c )/( ac + 2017c + 2017 ) = ( abc^2 )/( ac + abc^2 + abc ) = ( bc )/( 1 + bc + b )`

`⇒ a/( ab + a + 2017 ) + b/( bc + b + 1 ) + ( 2017 c )/( ac + 2017 c + 2017 ) = 1/( b + 1 + bc ) + ( bc )/( 1 + bc + b ) + ( b )/( 1 + bc + b ) = ( 1 + bc + b)/( 1 + bc + b ) = 1`

`⇒ P = 1`

Vậy `, P = 1 .`

Bài `2 :`

`a) ( 3x - 40 )/50 + ( 3x - 10 )/40 + ( x + 30 )/20 + ( x - 90 )/10 = 0`

`⇔ ( 12x - 160 )/200 + ( 15x - 50 )/200 + ( 10x + 300 )/200 + ( 20x - 1800 )/200 = 0`

`⇔ ( 12x - 160 + 15x - 50 + 10x + 300 + 20x - 1800 )/200 = 0`

`⇔ 12x - 160 + 15x - 50 + 10x + 300 + 20x - 1800 = 0`

`⇔ ( 12x + 15x + 10x + 20x ) - ( 160 + 50 - 300 + 1800 ) = 0`

`⇔ 57x - 1710 = 0`

`⇔ 57x = 1710`

`⇔ x = 30`

Vậy `,` tập nghiệm `S = { 30 } .`

b,

$P=\dfrac{a}{ab+a+2017}+\dfrac{b}{bc+b+1}+\dfrac{2017c}{ac+2017c+2017}$

$=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{ac+abc^2+abc}$

$=\dfrac{1}{b+bc+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{1+bc+b}$

$=\dfrac{b+bc+1}{b+bc+1}$

$=1$

Câu 2.

a,

$\dfrac{3x-40}{50}+\dfrac{3x-10}{40}+\dfrac{x+30}{20}+\dfrac{x-90}{10}=0$

$<=> \dfrac{3x-40-50}{50}+\dfrac{3x-10-80}{40}+\dfrac{x+30-60}{20}+\dfrac{x-90+60}{10}=0$

$<=>\dfrac{3(x-30)}{50}+\dfrac{3(x-30)}{40}+\dfrac{x-30}{30}+\dfrac{x-30}{10}=0$

$<=>(x-30)(\dfrac{3}{50}+\dfrac{3}{40}+\dfrac{1}{20}+\dfrac{1}{10})=0$

$<=>x-30=0$(Vế sau luôn dương)

$<=>x=30$

Vậy pt có nghiệm duy nhất $x=30$