Question

tìm y s/c giá trị của 2 biểu thức y+5/y-1 - y+1/y-3 và -8/(y-1)(y-3) bằng nhau

Answer 1

\(\dfrac{y+5}{y-1}-\dfrac{y+1}{y-3}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\) ( y # 1 ; y # 3)

⇔\(\dfrac{\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)}{\left(y-1\right)\left(y-3\right)}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)

⇔ ( y + 5)( y - 3) - ( y + 1)( y - 1) = - 8

⇔ y² - 3y + 5y - 15 - y² + 1 + 8 = 0

⇔ 2y - 6 = 0

⇔ y = 3 ( không thỏa mãn ĐKXĐ )

Vậy , phương trình vô nghiệm

Answer 2

\(\dfrac{y+5}{y-1}-\dfrac{y+1}{y-3}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)

\(\Leftrightarrow\dfrac{\left(y+5\right)\left(y-3\right)-\left(y-1\right)\left(y+1\right)}{\left(y-1\right)\left(y-3\right)}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)

\(\Leftrightarrow\dfrac{y^2+2y-15-y^2+1}{\left(y-1\right)\left(y-3\right)}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)

\(\Leftrightarrow2y-14=-8\)

\(\Leftrightarrow2y=6\)

\(\Leftrightarrow y=3\)

Vậy y = 3 thì .............