Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*)suy ra:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)\(=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
b) Tương tự câu a nhé bạn!
Câu b giải chi tiết như sau nhé:
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Từ đó, ta suy ra:
\(\dfrac{\left(a-b\right)^4}{\left(c-d\right)^4}=\dfrac{\left(bk-b\right)^4}{\left(dk-d\right)^4}=\dfrac{\left[b\left(k-1\right)\right]^4}{\left[d\left(k-1\right)\right]^4}=\dfrac{b^4.\left(k-1\right)^4}{d^4.\left(k-1\right)^4}=\dfrac{b^4}{d^4}\)(1)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\dfrac{b^4.k^4+b^4}{d^4.k^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\)
(2)
Từ (1) và (2) suy ra: \(\dfrac{\left(a-b\right)^4}{\left(c-d\right)^4}=\dfrac{a^4+b^4}{c^4+d^4}\)
