Question

[Phạm Thị Vân Anh]

Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

Answer 1

[Nguyễn Nguyệt Hằng]

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

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

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

Từ (1) và (2)=> \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)

Answer 2

[Nịna Hatori]

Dặt a/b = c/d = k

=> a = kb; c = kd

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