Question

cho tỉ lệ thức a/b=c/d chứng minh (a+b/c+d)^3 = a^3+b^3/c^3+d^3

Answer 1

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:\(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\left(\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}\right)^3=\dfrac{b^3}{d^3}\)(1)

Lại có :\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3.\left(k^3+1\right)}{d^3.\left(k^3+1\right)}=\dfrac{b^3}{d^3}\)(2)

Từ (1) và (2) => ĐPCM

Answer 2

Từ a/b=c/d

=>a/c=b/d=a+b/c+d

<=>a^3/c^3=b^3/d^3=(a+b)^3(c+d)^3

=a^3+b^3/c^3+d^3

Vậy

(a+b)^3(c+d)^3=a^3+b^3/c^3+d^3 (đpcm)