Question

\(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) (a,b,c,d khác 0)Chứng minh

\(\dfrac{a^2+b^2}{c^2+d^2}\) = \(\dfrac{\left(a-b^{ }\right)^2}{\left(c-d\right)^2}\)

Answer 1

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\left(1\right)\)

Thay (1) vào:

\(VT=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

\(VP=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2.\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}=VT\)

\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

-> ĐPCM.

Answer 2

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\)

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left(bk-b\right)\left(bk-b\right)}{\left(dk-d\right)\left(dk-d\right)}=\dfrac{\left[b\left(k-1\right)\right]\left[b\left(k-1\right)\right]}{\left[d\left(k-1\right)\right]\left[d\left(k-1\right)\right]}=\dfrac{b^2\left(k-1\right)}{d^2\left(k-1\right)}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\)\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

\(\rightarrowđpcm\)