Toán Lớp 8: Tìm a, b, c thoả mãn: $\dfrac{x^3}{x^4 – 1}$ = $\dfrac{a}{x – 1}$ + $\dfrac{b}{x + 1}$ + $\dfrac{cx + d}{x^2 + 1}$
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Toán Lớp 8: Tìm a, b, c thoả mãn: $\dfrac{x^3}{x^4 – 1}$ = $\dfrac{a}{x – 1}$ + $\dfrac{b}{x + 1}$ + $\dfrac{cx + d}{x^2 + 1}$
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ĐKXĐ : x\ne ±1
a/(x-1)+b/(x+1)+(cx+d)/(x^2+1)
=(a(x+1)+b(x-1))/(x^2-1)+(cx+d)/(x^2+1)
= (ax+ bx + a-b)/(x^2-1) + (cx+d)/(x^2+1)
= ((x^2+1)(ax+bx+a-b) + (cx+d)(x^2-1))/(x^4-1)
= (ax^3 + bx^3 + ax^2 – bx^2 + ax + bx +a-b + cx^3 – cx + dx^2-d)/(x^4-1)
= (x^3 (a+b+c) + x^2 (a-b +d)+ x (a+b -c) + (a-b-d))/(x^4-1)
-> x^3/(x^4-1) = (x^3(a+b+c)+x^2(a-b+d)+x(a+b-c)+(a-b-d))/(x^4-1)
-> x^3 = x^3(a+b+c) + (a-b+d)x^2 + (a+b-c)x+(a-b-d)
Vậy ta có 2 đa thức đồng nhất :
x^3 ≡x^3(a+b+c) + (a-b+d)x^2 + (a+b-c)x+(a-b-d)
<=> a+b+c=1 (1), a-b+d=0(2), a+b-c =0(3),a-b-d=0(4)
(1)+(2) -> 2a + c+d=1
->2a+c+d +a+b-c=1
-> 3a + b+d=1
->3a+b+d+a-b-d=1
-> 4a=1->a=1/4
(1) -> b+c=3/4
(2) -> b-d=1/4
(4) -> b+d=1/4
-> 2b = 2/4
->b=1/4
->c=1/2
->d=0
Vậy (a;b;c;d)=(1/4;1/4; 1/2;0)