Question

[你混過 vulnerable 他 難怪歐長]

1.Cho biểu thức M=\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10}{1\cdot6x+9x^2}\)

a, Tìm đk xác định của M

b, Rút gọn M

c, Tính giá trị của M vs x=\(\dfrac{1}{3}\)

2. CMR M k phụ thuộc vào x : \(P=\left(\dfrac{x}{x^2-25}-\dfrac{x-5}{x^2+5x}\right):\dfrac{2x-5}{^{ }x^2+5x}^{ }+\dfrac{x}{5-x}\)

Cần gấp . Thanks

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

a: ĐKXĐ: x<>1/3; x<>-1/3

b: \(M=\left(\dfrac{-3x}{3x-1}+\dfrac{2x}{3x+1}\right)\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)

\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)

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

Bài 2:

\(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)

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

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