Đáp án + Giải thích các bước giải:
Bài `3` :
`a)` ĐKXĐ : `{(5x+25ne0),(xne0),(x^2+5xne0):}<=>{(xne-5),(xne0),(x(x+5)ne0):}<=>{(xne-5),(xne0),(xne0),(xne-5):}`
`<=>xne0,xne-5`
`b)` `Q=x^2/(5x+25)+(2x-10)/x+(50+5x)/(x^2+5x)(xne0,xne-5)`
`=x^2/[5(x+5)]+(2x-10)/x+(50+5x)/[x(x+5)]`
`=(x^2*x)/[5x(x+5)]+[(2x-10)*5(x+5)]/[5x(x+5)]+[(50+5x)*5]/[5x(x+5)]`
`=[x^3+(2x-10)(5x+25)+250+25x]/[5x(x+5)]`
`=(x^3+10x^2+50x-50x-250+250+25x)/[5x(x+5)]`
`=(x^3+10x^2+25x)/[5x(x+5)]`
`=[x(x^2+10x+25)]/[5x(x+5)]`
`=[x(x+5)^2]/[5x(x+5)]=(x+5)/5`
`c)` `Q=-4<=>(x+5)/5=-4`
`<=>(x+5)/5=-20/5`
`=>x+5=-20<=>x=-25`
Vậy `x = -25` để `P = -4`
Bài `2` :
`a)` ĐKXĐ : `x^2-5xne0<=>x(x-5)ne0<=>[(xne0),(xne5):}`
`b)` `P=(x^2-10x+25)/(x^2-5x)=[(x-5)^2]/[x(x-5)]=(x-5)/x`
`P=5/2<=>(x-5)/x=2`
`<=>(x-5)/x=(2x)/x`
`=>x-5=2x`
`<=>x-5-2x=0`
`<=>-x-5=0`
`<=>-x=5`
`<=>x=-5`(TM)
`c)` `(x-5)/x=1-5/x`
`5vdotsx<=>x in Ư(5)={1;-1;5;-5}`
Mà `xne0,xne5` nên ta loại bỏ `x = 5`
Vậy `x in {1;-1;-5}`.
