Question

[Oh my darling❤]

Kíu típ :__________(  

Answer 1

[乇尺尺のﾚ]

\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

⇔\(\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4+1+1+1+1\)

⇔\(\dfrac{x+1}{99}+\dfrac{99}{99}+\dfrac{x+2}{98}+\dfrac{98}{98}+\dfrac{x+3}{97}+\dfrac{97}{97}+\dfrac{x+4}{96}+\dfrac{96}{96}=-4+4\)

⇔\(\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

⇔\(\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

⇔\(x+100=0\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\ne0\right)\)

⇔\(x=-100\)

cíu được phần 1 thôi nhé

Answer 2

[Thuỳ Linh Nguyễn]

Bổ xung ý 2

\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\\ \Rightarrow\dfrac{1}{x}=\dfrac{5}{6}-\dfrac{y}{3}\\ \Rightarrow\dfrac{1}{x}=\dfrac{5-2y}{6}\\ \Rightarrow x\cdot\left(5-2y\right)=6\)

`=>x;5-2y in Ư(6)={+-1;+-3;+-2;+-6}`

mà `5-2y` là số lẻ 

nên `5-2y in {+-1;+-3}`

Ta có bảng sau :

`5-2y`	`-1`	`-3`	`1`	`3`
`y`	`3(T//m)`	`4(T//m)`	`2(T//m)`	`1(T//m)`
`x`	`-1(L)`	`-3(L)`	`1(T//m)`	`3(T//m)`

Vậy `x;y in {(1;2);(3;1)}`