Question

[Thanh Thủy Nguyễn]

em đang cần gấp ạ, em cảm ơn ( bài 1,2)

Answer 1

[Nguyễn Việt Lâm]

1.

a. ĐKXĐ: \(x\ne\left\{0;-5;5\right\}\)

\(A=\left(\dfrac{x}{x+5}+\dfrac{5}{x-5}+\dfrac{10x}{x^2-25}\right).\left(1-\dfrac{5}{x}\right)\)

\(=\left(\dfrac{x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{10x}{\left(x-5\right)\left(x+5\right)}\right).\left(\dfrac{x-5}{x}\right)\)

\(=\left(\dfrac{x^2-5x+5x+25+10x}{\left(x-5\right)\left(x+5\right)}\right)\left(\dfrac{x-5}{x}\right)\)

\(=\left(\dfrac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}\right)\left(\dfrac{x-5}{x}\right)\)

\(=\dfrac{\left(x+5\right)^2\left(x-5\right)}{\left(x-5\right)\left(x+5\right).x}=\dfrac{x+5}{x}\)

b.

\(A=\dfrac{x+5}{x}=1+\dfrac{5}{x}\)

\(A\in Z\Rightarrow\dfrac{5}{x}\in Z\Rightarrow5⋮x\)

\(\Rightarrow x=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

Thay vào ta được \(A=\left\{0;-4;6;2\right\}\)

Do A nguyên dương nên loại \(A=\left\{0;-4\right\}\) tương ứng \(x=\left\{-5;-1\right\}\)

Vậy \(x=\left\{1;5\right\}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

2:

=>4=(2x-5)(x-1)-2x(x+3)

=>2x^2-2x-5x+5-2x^2-6x=4

=>-13x+5=4

=>-13x=-1

=>x=1/13