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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: ( $\frac{21}{x^2-9}$ – $\frac{x-4}{3-x}$ – $\frac{x-1}{3+x}$ ):(1- $\frac{1}{x+3}$ )

Tháng Tám 26, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: ( $\frac{21}{x^2-9}$ – $\frac{x-4}{3-x}$ – $\frac{x-1}{3+x}$ ):(1- $\frac{1}{x+3}$ )

Comments ( 2 )

  1. bichhhathu
    bichhhathu
    26 Tháng Tám, 2022 at 4:43 sáng
    Reply
    ĐKXĐ:x\ne+-3
    (21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))
    =[21/((x+3)(x-3))+(x-4)/(x-3)-(x+1)/(x+3)]:(x+3-1)/(x+3)
    =(21+(x-4)(x-3)-(x-1)(x-3))/((x+3)(x-3)) . (x+3)/(x+2)
    =(21+x^2-x-12-(x^2-4x+3))/((x+2)(x-3))
    =(21+x^2-x-12-x^2+4x-3)/((x+2)(x+3))
    =(3x+6)/((x+2)(x+3))
    =(3(x+2))/((x+3)(x+2))
    =3/(x-3)
     

  2. thienthanh
    thienthanh
    26 Tháng Tám, 2022 at 4:43 sáng
    Reply
    Giải đáp:
    (21/(x^(2)-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))   (ĐK 😡 \ne+-3)
    =(21/(x^(2)-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))
    =(21/[(x-3)(x+3)]+[(x-4)(x+3)]/[(x+3)(x-3)]-[(x-1)(x-3)]/[(x+3)(x-3)]):((x+2)/(x+3))
    =((21+x^(2)+3x-4x-12-x^(2)+3x+x-3)/[(x+3)(x-3)]):((x+2)/(x+3))
    =((3x+6)/[(x+3)(x-3)]):((x+2)/(x+3))
    =[3(x+2)(x+3)]/[(x+3)(x-3).(x+2)]
    =3/(x-3)
     

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222-9+11+12:2*14+14 = ? ( )

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