Giải thích các bước giải:
a.Ta có:
$\dfrac{y+1}{2y^2-7y+3}+\dfrac{4+y}{2y^2-5y+2}=\dfrac{5+2y}{2y^2-7y+3}$
$\to \dfrac{4+y}{2y^2-5y+2}=\dfrac{5+2y}{2y^2-7y+3}-\dfrac{y+1}{2y^2-7y+3}$
$\to \dfrac{4+y}{2y^2-5y+2}=\dfrac{5+2y-(y+1)}{2y^2-7y+3}$
$\to \dfrac{y+4}{2y^2-5y+2}=\dfrac{y+4}{2y^2-7y+3}$
$\to y+4=0\to y=-4$
Hoặc $2y^2-5y+2=2y^2-7y+3(\ne 0)$
$\to 2y=1\to y=\dfrac12$ loại vì khi $x=\dfrac12\to 2y^2-5y+2=0$
Vậy $y=-4$
b.Ta có:
$1-\dfrac{2b}{x-b}=\dfrac{a-b}{b^2+x^2-2bx}$
$\to 1-\dfrac{2b}{x-b}=\dfrac{a-b}{(x-b)^2}$
$\to (x-b)^2-2b(x-b)=a-b$
$\to (x-b)^2-2b(x-b)+b^2=a-b+b^2$
$\to (x-b-b)^2=a-b+b^2$
$\to (x-2b)^2=a-b+b^2$
$\to x-2b=\pm\sqrt{a-b+b^2}$
$\to x=2b\pm\sqrt{a-b+b^2}$
