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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: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Ta có: \(b^2=ac\)
=>\(\frac{a}{b}=\frac{b}{c}\) (1)
\(c^2=bd\)
=>\(\frac{b}{c}=\frac{c}{d}\) (2)
Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
=>c=dk; \(b=ck=dk\cdot k=dk^2\) và \(a=bk=dk^2\cdot k=dk^3\)
\(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}=\frac{3\cdot\left(dk^3\right)^3-4\cdot\left(dk^2\right)^3+5\cdot\left(dk\right)^3}{3\cdot\left(dk^2\right)^3-4\cdot\left(dk\right)^3+5\cdot d^3}\)
\(=k^3\cdot\frac{3\cdot d^3\cdot k^6-4\cdot d^3\cdot k^3+5d^3}{3d^3\cdot k^6-4\cdot d^3\cdot k^3+5\cdot d^3}=k^3\)
\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)
Do đó: \(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}=\frac{a}{d}\)
\(Cho \dfrac{a}{b}=\dfrac{c}{d} ;b+d khác 0 CM \dfrac{3a^2+c^2}{3b^2+d^2}=\dfrac{(a+c)^2}{(b+d)^2}\)
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...