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

Toán Lớp 8: $\frac{x^2+9}{x^2+3x}$ +$\frac{6}{x+3}$

Tháng Tư 9, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: $\frac{x^2+9}{x^2+3x}$ +$\frac{6}{x+3}$

Comments ( 2 )

  1. khanhlanchau
    khanhlanchau
    9 Tháng Tư, 2022 at 9:23 sáng
    Reply
    (x^2+9)/(x^2+3x)+6/(x+3)(x\ne0;x\ne-3)
    =(x^2+9)/(x(x+3))+(6x)/(x(x+3))
    =(x^2+9+6x)/(x(x+3))
    =((x+3)^2)/(x(x+3))
    =(x+3)/x

  2. hienthucthu97
    hienthucthu97
    9 Tháng Tư, 2022 at 9:23 sáng
    Reply
    \frac{ x^2+9}{x^2+3x} + \frac{6}{x+3}            ( x \ne 0 ; x \ne -3 )
    = \frac{x^2+9}{x ( x + 3 )} + \frac{6}{x+3}
    = \frac{x^2+9+6x}{x ( x + 3 ) }
    = \frac{ ( x + 3 )^2}{x ( x + 3 )}
    = \frac{ x+3 }{x}

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

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