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

Toán Lớp 8: Tìm đa thức A thỏa mãn điểu kiện: $a)$ $\frac{A(x-5)}{x^{2}-4x-5}=$ $\frac{3x^{2}+9x}{x^{2}+4x+3}$ $b)$ $\frac{x^{2}+x-6}{A(x+3)}=$ $

Home/ Toán Lớp 8: Tìm đa thức A thỏa mãn điểu kiện: $a)$ $\frac{A(x-5)}{x^{2}-4x-5}=$ $\frac{3x^{2}+9x}{x^{2}+4x+3}$ $b)$ $\frac{x^{2}+x-6}{A(x+3)}=$ $

Toán Lớp 8: Tìm đa thức A thỏa mãn điểu kiện: $a)$ $\frac{A(x-5)}{x^{2}-4x-5}=$ $\frac{3x^{2}+9x}{x^{2}+4x+3}$ $b)$ $\frac{x^{2}+x-6}{A(x+3)}=$ $

Tháng Ba 2, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm đa thức A thỏa mãn điểu kiện:
$a)$ $\frac{A(x-5)}{x^{2}-4x-5}=$ $\frac{3x^{2}+9x}{x^{2}+4x+3}$
$b)$ $\frac{x^{2}+x-6}{A(x+3)}=$ $\frac{(5x-1)(5x-2)}{5x^{3}-x^{2}+15x-3}$

Comments ( 2 )

  1. dinhcongson
    dinhcongson
    2 Tháng Ba, 2023 at 5:43 chiều
    Reply
    Gửi bạn:
    $a,ĐKXĐ:x\neq 5,-5,-3$
    $\dfrac{A(x-5)}{x^2-4x-5}=\dfrac{3x^2+9x}{x^2+4x+3}$ 
    $\dfrac{A(x-5)}{x^2+x-5x-5}=\dfrac{3x(x+3)}{x^2+3x+x+3}$ 
    $\dfrac{A(x-5)}{x(x+1)-5(x+1)}=\dfrac{3x(x+3)}{x(x+3)+x+3}$ 
    $\dfrac{A(x-5)}{(x-5)(x+1)}=\dfrac{3x(x+3)}{(x+3)(x+1)}$ 
    $\dfrac{A}{x+1}=\dfrac{3x}{x+1}$ 
    $⇒A=3x$
    $b,ĐKXĐ: x\neq -3,\dfrac{1}{4},\dfrac{2}{5}$
    $\dfrac{x^2+x-6}{A(x+3)}=\dfrac{(5x-1)(5x-2)}{5x^3-x^2+15x-3}$ 
    $\dfrac{x^2+3x-2x-6}{A(x+3)}=\dfrac{(5x-1)(5x-2)}{x^2(5x-1)+3(5x-1)}$ 
    $\dfrac{x(x+3)-2(x+3)}{A(x+3)}=\dfrac{(5x-1)(5x-2)}{x^2(5x-1)+3(5x-1)}$ 
    $\dfrac{x-2}{A}=\dfrac{5x-2}{x^2+3}$ 
    $A.(5x-2)=(x-2)(x^2+3)$ 
    $A=\dfrac{(x-2)(x^2+3)}{5x-2}$ 

  2. diemmyhoa
    diemmyhoa
    2 Tháng Ba, 2023 at 5:42 chiều
    Reply
    Giải đáp:
    a, {A.(x – 5)}/{x^2 – 4x – 5} = {3x^2 + 9x}/{x^2 + 4x + 3}
    Ta có:
    {A.(x – 5)}/{x^2 – 4x – 5}
    = {A.(x – 5)}/{x^2 + x – 5x – 5}
    = {A.(x – 5)}/{x.(x + 1) – 5.(x + 1)}
    = {A.(x – 5)}/{(x + 1).(x – 5)}
    = A/{x + 1}
    **
    {3x^2 + 9x}/{x^2 + 4x + 3}
    = {3x.(x + 3)}/{x^2 + x + 3x + 3}
    = {3x.(x + 3)}/{x.(x + 1) + 3.(x + 1)}
    = {3x.(x + 3)}/{(x + 1).(x + 3)}
    = {3x}/{x + 1}
    =>
    A/{x + 1} = {3x}/{x + 1}
    -> A = 3x
    Vậy A = 3x
    ———————-
    b, {x^2 + x – 6}/{A.(x + 3)} = {(5x – 1).(5x – 2)}/{5x^3- x^2 + 15x – 3}
    Ta có:
    {x^2 + x – 6}/{A.(x + 3)}
    = {x^2 + 3x – 2x – 6}/{A.(x + 3)}
    = {x.(x + 3) – 2.(x + 3)}/{A.(x + 3)}
    = {(x + 3).(x – 2)}/{A.(x + 3)}
    = {x – 2}/A
    **
    {(5x – 1).(5x – 2)}/{5x^3- x^2 + 15x – 3}
    = {(5x – 1).(5x – 2)}/{x^2.(5x – 1) + 3.(5x – 1)}
    = {(5x – 1).(5x – 2)}/{(5x – 1).(x^2 + 3)}
    = {5x – 2}/{x^2 + 3}
    =>
    {x – 2}/A = {5x – 2}/{x^2 + 3}
    -> (x^2 + 3).(x – 2) = (5x – 2).A
    -> A = [(x^2 + 3).(x – 2)] : (5x – 2)
    -> A = ((x^2 + 3).(x – 2))/(5x – 2)
    Vậy A = ((x^2 + 3).(x – 2))/(5x – 2)
    #dariana

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

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