Question

[Hùng Chu]

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

Answer 1

[Nguyễn Nho Bảo Trí]

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

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

dkxd : x ≠ 0

          x ≠ 5

          x ≠ -5

MTC : 2x(x - 5)(x + 5)

Quy đồng mẫu thức hai vế của phương trình :

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

Suy ra : 2(x - 5)(x + 5) - (x - 5)(x + 5) = x(x + 25)

         \(\Leftrightarrow\) 2(x² - 25) - (x² - 25) = x² + 25x

         \(\Leftrightarrow\) 2x² - 50 - x² + 25 - x² - 25x = 0

        \(\Leftrightarrow\) -25 - 25x = 0

        \(\Leftrightarrow\) -25x = 25

        \(\Leftrightarrow\) x = \(\dfrac{25}{-25}=-1\) (thỏa mãn)

 Vậy S = \(\left\{-1\right\}\)

 Chúc bạn học tốt

Answer 2

[Nguyễn Lê Phước Thịnh]

Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

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

Suy ra: \(2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)

\(\Leftrightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

hay \(x=-\dfrac{5}{3}\)(thỏa ĐK)