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 \(x=\left\{1;5\right\}\)
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
2.
ĐKXĐ: \(x\ne\left\{-3;1\right\}\)
\(\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}+\dfrac{2x}{1-x}\)
\(\Leftrightarrow\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}-\dfrac{2x}{x-1}\)
\(\Leftrightarrow\dfrac{4}{\left(x+3\right)\left(x-1\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\dfrac{2x\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}\)
\(\Rightarrow4=\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)\)
\(\Leftrightarrow4=2x^2-7x+5-\left(2x^2+6x\right)\)
\(\Leftrightarrow-13x=-1\)
\(\Leftrightarrow x=\dfrac{1}{13}\) (thỏa mãn)
mn giải giúp em vơi ạ em đang cần gấp cảm ơn
giúp em bài 1 dạng 3 vói em, em cần gấp lắm, em cảm ơn ạ
cảm ơn mọi người đã chỉ em đang cần gấp ạ câu 2,3,4,5 làm mỗi 4 câu đó thôi ạ