Ai giúp em với ạ!!!!!!!!!!!!!!!!!!!!!!!!!

`a,P = (x^2+2x)/(2x+10)+(x-5)/x + (50-5x)/(2x(x+5)) (ĐKX Đ : x\ne -5;0)`
`=>P = (x^2 + 2x)/(2(x+5)) + (x-5)/x + (50-5x)/(2x(x+5))`
`=>P = (x(x^2 + 2x) + 2(x+5)(x-5)+50-5x)/(2x(x+5))`
`=>P = (x^3 + 2x^2 + 2x^2 - 50 + 50-5x)/(2x(x+5))`
`=>P = (x^3+4x^2-5x)/(2x(x+5)) = (x(x^2 + 4x - 5))/(2x(x+5))`
`=>P = (x^2-x+5x-5)/(2(x+5)) = (x(x-1)+5(x-1))/(2(x+5))`
`=>P = ((x+5)(x-1))/(2(x+5)) = (x-1)/2`
Vậy `P=(x-1)/2`
`b,`Để `P=1/4`
`<=>(x-1)/2 = 1/4`
`<=>4(x-1) = 2`
`<=>4x - 4 = 2`
`<=>4x=6`
`<=>x=3/2`
Vậy `x=3/2` tại `P=1/4`
`c,`Để `P<0`
`<=>(x-1)/2 < 0`
`<=>x-1<0`
`<=>x < 1` hay \(\left[ \begin{array}{l}x<1\\x\ne0\\x\ne-5\end{array} \right.\) 
`d,Q =(x-1)/2 . (2x-6)`
`=>Q = ((x-1)(2x-6))/2 = (x-1)(x-3) = x^2 - 3x - x+3 = x^2 - 4x + 2 + 1 =(x-2)^2 - 1 >=-1AAx`
Dấu `'='` xảy ra `<=>(x-2)^2 = 0`
`<=>x-2=0`
`<=>x=2`
Vậy `Q_{min} = -1 <=>x=2`