Đáp án+Giải thích các bước giải:
Bài 4:
`a) B=(x/(x+3)+x(x-3)-2/(x^2-9)).(x+3)/(2x-2)(ĐKXĐ: x \ne±3; x \ne 1)`
`=(x(x-3)/(x^2-9)+(x(x+3))/(x^2-9)-2/(x^2-9)).(x+3)/(2x-2)`
`= (x^2-3x+x^2+3x-2)/(x^2-9).(x+3)/(2(x-1))`
`= (2x^2-2)/((x+3)(x-3)).(x+3)/(2(x-1))`
`= (2(x-1)(x+1))/((x+3)(x-3)).(x+3)/(2(x-1))`
`=(x+1)/(x-3)`
`b)` Để `B=2` thì `(x+1)/(x-3)=2`
`<=> x+1 =2(x-3)`
`<=> x+1=2x-6`
`<=> -x = -7`
`<=> x =7`
Bài 5:
`a)P= ((21)/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3)) (ĐKXĐ: x \ne ±3; x \ne 2)`
`= ((21)/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3)/(x+3)-1/(x+3))`
`=((21)/(x^2-9)+((x-4)(x+3))/(x^2-9)-((x-1)(x-3))/(x^2-9)):(x+3-1)/(x+3)`
`= (21+x^2+3x-4x-12-x^2+3x+x-3)/(x^2-9):(x+2)/(x+3)`
`= (3x+6)/(x^2-9):(x+2)/(x+3)`
`= (3(x+2))/((x-3)(x+3)).(x+3)/(x+2)`
`=3/(x-3)`
`b)` Ta có: `x^2=3x`
`<=> x^2-3x=0`
`<=> x(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=0(t/m)\\x=3(Loại)\end{array} \right.\)
Với `x=0` thì `P` có giá trị là:
`P=3/(0-3)=3/(-3)=-1`
