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

Toán Lớp 8: f) $\frac{x^2}{x^3 – 4x}$ + $\frac{6}{6 – 3x}$ + $\frac{1}{x+2}$ g) $\frac{5}{x – 2}$ – $\frac{1}{x + 2}$ + $\frac{4}{x^2 – 4}$

Toán Lớp 8: f) $\frac{x^2}{x^3 – 4x}$ + $\frac{6}{6 – 3x}$ + $\frac{1}{x+2}$
g) $\frac{5}{x – 2}$ – $\frac{1}{x + 2}$ + $\frac{4}{x^2 – 4}$

Comments ( 2 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
     f) x^2(x^3-4x)+6/(6-3x)+1/(x+2) (Điều kiện xác định: x \ne 0; +-2)
    =x^2/[x(x^2-4)]+6/[3(2-x)]+1/(x+2)
    =x^2/[x(x+2)(x-2)]-6/[3(x-2)]+1/(x+2)
    =x/[(x+2)(x-2)]-2/(x-2)+1/(x+2)
    Mẫu thức chung: (x+2)(x-2)
    =x/[(x+2)(x-2)]-[2(x+2)]/[(x+2)(x-2)]+(x-2)/[(x+2)(x-2)]
    =x/[(x+2)(x-2)]-(2x+4)/[(x+2)(x-2)]+(x-2)/[(x+2)(x-2)]
    =(x-2x-4+x-2)/[(x+2)(x-2)]
    =(-6)/[(x+2)(x-2)]
    g) 5/(x-2)-1/(x+2)+4/(x^2-4) (điều kiện xác đinh: x \ne +-2)
    =5/(x-2)-1/(x+2)+4/[(x+2)(x-2)]
    Mẫu thức chung: (x+2)(x-2)
    =[5(x+2)]/[(x+2)(x-2)]-(x-2)/[(x+2)(x-2)]+4/[(x+2)(x-2)]
    =(5x+10)/[(x+2)(x-2)]-(x-2)/[(x+2)(x-2)]+4/[(x+2)(x-2)]
    =(5x+10-x+2+4)/[(x+2)(x-2)]
    =(4x+16)/[(x+2)(x-2)]

  2. f)x^2 / {x^3-4x}+6 / {6-3x} +1 / {x+2}    (x\ne0;x\ne+-2)
    =x^2 / {x(x^2-4)}+6 / {3(2-x)} +1 / {x+2}
    =x/ {x^2-4}-2 / {x-2} +1 / {x+2}
    =(x-2(x+2)+x-2)/((x+2)(x-2))
    =(x-2x-4+x-2)/((x+2)(x-2))
    =(-6)/((x-2)(x+2))
    g)5/(x-2)-1/(x+2)+4/(x^2-4)   (x\ne+-2)
    =[5(x+2)-(x-2)+4]/[(x-2)(x+2)]
    =[5x+10-x+2+4]/[(x-2)(x+2)]
    =[4x+16]/[(x-2)(x+2)]
     

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

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