Toán Lớp 7: Cho a/b=c/d. Chứng minh rằng 3a^5+c^5/3b^5+d^5=(a+c)^5/(b+d)^5
giúp mình với ^^
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Toán Lớp 7: Cho a/b=c/d. Chứng minh rằng 3a^5+c^5/3b^5+d^5=(a+c)^5/(b+d)^5
giúp mình với ^^
Comments ( 2 )
Giải đáp:
Lời giải và giải thích chi tiết:
Đặt a/b =c/d = k ( k \ne 0 ) . Suy ra $\begin{cases} a = kb \\ c = kd \end{cases}$
Khi đó , ta được :
(3a^5+c^5)/(3b^5+d^5) = (3k^5b^5 + k^5d^5)/(3b^5 + d^5) = (k^5(3b^5 +d^5))/(3b^5 +d^5) = k^5
(a+c)^5/(b+d)^5 = (kb +kd)^5/(b+d)^5= ( [k(b+d)]^5)/(b+d)^5 = (k^5(b+d)^5)/(b+d)^5 = k^5
=> (3a^5+c^5)/(3b^5+d^5)=(a+c)^5/(b+d)^5( =k^5)
Đặt a/b=c/d=k(k\ne 0)
->a=bk,c=dk
(3a^5+c^5)/(3b^5+d^5)
=(3b^5k^5 +d^5k^5)/(3b^5+d^5)
=(k^5 (3b^5 + d^5))/(3b^5+d^5)
=k^5(1)
(a+c)^5/(b+d)^5
=(bk+dk)^5/(b+d)^5
=(k(b+d))^5/(b+d)^5
=(k^5(b+d)^5)/(b+d)^5
=k^5(2)
(1)(2)->(3a^5+c^5)/(3b^5+d^5)=(a+c)^5/(b+d)^5