Đáp án + Giải thích các bước giải:
`a)` `(x -3)^2+2x-6=0` $\\$ `<=> (x-3)^2+2(x-3)=0` $\\$ `<=> (x-3)(x-3+2)=0` $\\$ `<=> (x-3)(x-1) = 0 <=> `\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
Vậy `S = {3;1}`
`b)` `(x + 3)/(x - 3)+48/(9-x^2)=(x-3)/(x+3)(x ne pm3)` $\\$ `<=> (x+3)/(x-3)-48/[(x-3)(x+3)] = (x-3)/(x+3)` $\\$ `<=> [(x+3)(x+3)]/[(x-3)(x+3)] - 48/[(x-3)(x+3)] = [(x-3)(x-3)]/[(x-3)(x+3)]` $\\$ `=> (x+3)(x+3) - 48 = (x-3)(x-3)` $\\$ `<=> x^2 + 6x + 9 -48 = x^2 - 6x + 9` $\\$ `<=> x^2 + 6x + 9 - 48 - x^2 + 6x - 9 = 0` $\\$ `<=> 12x -48 = 0 <=> x = 4(TM)`
Vậy `S = {4}`
`c)` TH1:
`x/2 + 1 - x/3>0` $\\$ `<=>6*x/2+1*6-6*x/3>0*6` $\\$ `<=> 3x+6-2x > 0` $\\$ `<=> x + 6 > 0 <=> x > -6`
Vậy `S = {x|x > -6}`
TH2 :
`x/2 + (1-x)/3 >0` $\\$ `<=> 6*x/2+6*(1-x)/3>0*6` $\\$ `<=> 3x + 2(1-x)>0` $\\$ `<=> 3x + 2 - 2x > 0` $\\$ `<=> x + 2 > 0 <=> x > -2`
Vậy `S = {x| x> -2}`
`d)` `(x - 9)^2 - x(x + 9) <=0` $\\$ `<=>x^2 - 18x+81-x^2-9x<=0` $\\$ `<=>-27x+81<=0` $\\$ `<=> -27x <= -81` $\\$ `<=> -27x:(-27) >= (-81):(-27)` $\\$ `<=> x>=3`
Vậy `S = {x|x>=3}`
