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

Toán Lớp 8: $\frac{y+8}{y^{2}-16}$ – $\frac{2}{y^{2}+4y}$

Toán Lớp 8: $\frac{y+8}{y^{2}-16}$ – $\frac{2}{y^{2}+4y}$

Comments ( 2 )

  1. (y+8)/(y^2 – 16)  – 2/(y^2 + 4y)
    = (y+8)/[(y- 4)(y+4)]  – 2/[y(y + 4)]
    = (y^2+8y)/[y(y- 4)(y+4)] – (2y – 8)/[y(y + 4)(y-4)]
    = (y^2+8y-2y + 8)/[y(y- 4)(y+4)]
    = (y^2+6y + 8)/[y(y- 4)(y+4)]

  2. Giải đáp+Lời giải và giải thích chi tiết:
    $\dfrac{y+8}{y^{2}-16}$-$\dfrac{2}{y^{2}+4y}$
    =$\dfrac{y+8}{y^{2}-4^{2}}$-$\dfrac{2}{y.y+y.4}$
    =$\dfrac{y+8}{(y-4)(y+4)}$-$\dfrac{2}{y(y+4)}$
    =$\dfrac{(y+8).y}{y(y+4)(y-4)}$-$\dfrac{2(y-4)}{y(y+4)(y-4)}$
    =$\dfrac{y^{2}+8y}{y(y+4)(y-4)}$-$\dfrac{2y-8}{y(y+4)(y-4)}$
    =$\dfrac{(y^2+8y)-(2y-8)}{y(y+4)(y-4)}$
    =$\dfrac{y^2+8y-2y+8}{y(y+4)(y-4)}$
    =$\dfrac{y^2+6y+8}{y(y+4)(y-4)}$
    ->Áp dụng HĐT: A^2-B^2=(A-B)(A+B)
    $\text{#LinhSad}$

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