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Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
\(Cho \dfrac{a}{b}=\dfrac{c}{d} ;b+d khác 0 CM \dfrac{3a^2+c^2}{3b^2+d^2}=\dfrac{(a+c)^2}{(b+d)^2}\)
Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{3a^2+c^2}{3b^2+d^2}=\dfrac{3b^2k^2+d^2k^2}{3b^2+d^2}=\dfrac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\) (1)
\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{3a^2+c^2}{3b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(đpcm\right)\)
Vậy...