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

Toán Lớp 8: `1/{x + 2} + 3/{x^2 – 4} + {x – 14}/{(x^2 + 4x + 4)(x – 2)}`

Home/ Toán Lớp 8: `1/{x + 2} + 3/{x^2 – 4} + {x – 14}/{(x^2 + 4x + 4)(x – 2)}`

Toán Lớp 8: `1/{x + 2} + 3/{x^2 – 4} + {x – 14}/{(x^2 + 4x + 4)(x – 2)}`

Tháng Bảy 11, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: 1/{x + 2} + 3/{x^2 – 4} + {x – 14}/{(x^2 + 4x + 4)(x – 2)}

Comments ( 2 )

  1. kymmailien
    kymmailien
    11 Tháng Bảy, 2022 at 2:25 chiều
    Reply
    Ta có: x^2 + 4x + 4 = x^2 + 2.2.x + 2^2 = (x + 2)^2 
    x^2-4 = ( x + 2 )( x – 2 )
    $\dfrac{1}{x + 2 }$ + $\dfrac{3}{x^2 – 4 }$ + $\dfrac{x – 14}{(x^2 + 4x + 4)(x – 2)}$
    = $\dfrac{( x + 2 )(x – 2) + 3(x+2) + x – 14}{(x^2 + 4x + 4)(x – 2) }$ 
    = $\dfrac{x^2 – 4 + 3x + 6 + x – 14}{(x^2 + 4x + 4)(x – 2)}$
    = $\dfrac{x^2 + 4x – 12}{(x^2 + 4x + 4)(x – 2)}$
    = $\dfrac{(x^2 + 4x + 4) – 16}{(x^2 + 4x + 4)(x – 2)}$
    = $\dfrac{(x + 2 – 4)(x + 2 + 4) }{(x^2 + 4x + 4)(x – 2)}$
    = $\dfrac{(x – 2 )(x + 6)}{(x^2 + 4x + 4)(x – 2)}$
    = $\dfrac{x + 6}{(x+2)^2}$

  2. baoquyen
    baoquyen
    11 Tháng Bảy, 2022 at 2:25 chiều
    Reply
    $\text{ @HV }$
    Có gì không hiểu bạn hỏi lại nhé!
     

    toan-lop-8-1-2-3-2-4-14-2-4-4-2

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

Kim Xuân

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