\(\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}=\frac{b-a}{ab}=\frac{1}{ab}\Rightarrow b-a=1\)

vậy với \(a;b\in Z\)sao cho b=a+1 thì \(\frac{1}{a}-\frac{1}{b}=\frac{1}{ab}\)

1/a-1/b=b-a/ab=1/ab

Do đó ab(a-b)=ab, nên b-a=ab:ab

=>b-a=1

=>b=a+1

a. \(A=\left[\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right].\frac{x+7}{x}\)

\(=\left[\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right].\frac{x+7}{x}\)

\(=\left[\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{x^2-1}\right].\frac{x+7}{x}\)

\(=\frac{x^2-1}{x^2-1}.\frac{x+7}{x}\)

\(=\frac{x+7}{x}\)

b. Để A \(\in\)Z thì \(\frac{x+7}{x}\in Z\)

=> x+7 chia hết cho x

Mà x chia hết cho x

=> 7 chia hết cho x

=> x \(\in\)Ư(7)={-7; -1; 1; 7}

Vậy x \(\in\){-7; -1; 1; 7} thì A \(\in\)Z.

Hoàng Bảo Ngọc trình bày cách làm cho tau với

ui mk bó tay vì chưa hok đến lóp 9!!! ^^

54746767765858578758788974686865876546456475675685785

a: \(2019\left|x-2019\right|+\left(x-2019\right)^2=2018\left|2019-x\right|\)

=>\(\left(x-2019\right)^2+2019\left|x-2019\right|-2018\left|x-2019\right|=0\)

=>\(\left(\left|x-2019\right|\right)^2+\left|x-2019\right|=0\)

=>|x-2019|(|x-2019|+1)=0

=>|x-2019|=0

=>x-2019=0

=>x=2019

b: Sửa đề: \(\frac{2x+1}{7}=\frac{1}{y}\)

=>y(2x+1)=7

mà 2x+1 lẻ

nên (2x+1;y)∈{(1;7);(7;1);(-1;-7);(-7;-1)}

=>(2x;y)∈{(0;7);(6;1);(-2;-7);(-8;-1)}

=>(x;y)∈{(0;7);(3;1);(-1;-7);(-4;-1)}

Cho mình sửa lại đề câu 1b: \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\frac{x}{7}-\frac{1}{2}=\frac{1}{y+1}\)

\(\frac{2x-7}{14}=\frac{1}{y+1}\)

\(TH1:\hept{\begin{cases}2x-7=7\\y+1=2\end{cases}\Rightarrow\hept{\begin{cases}x=7\\y=1\end{cases}}}\)

\(TH2:\hept{\begin{cases}2x-7=-7\\y+1=-2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}}\)

nhớ cho