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/b = c/d => a/c = b/d
=> a2 / c2 = b2 / d2 = ab / cd
<=> 7a2 / 7c2 = 11a2 / 11c2 = 8b2 / 8d2 = 3ab / 3cd
=> 7a2 + 3ab / 7c2 + 3cd = 11a2 - 8b2 / 11c2 - 8d2
=> 7a2 + 3ab / 11a2 - 8b2 = 7c2 + 3cd / 11c2 - 8d2
=> (đpcm)
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ (1) và (2), suy ra: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
Vậy \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
~ Học tốt!~
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) CMR:
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2t^2+3b^2t}{11b^2t^2-8b^2}=\dfrac{b^2\left(7t^2+3t\right)}{b^2\left(11t^2-8\right)}=\dfrac{7t^2+3t}{11t^2-8}\\\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2t^2+3d^2t}{11d^2t^2-8d^2}=\dfrac{d^2\left(7t^2+3t\right)}{d^2\left(11t^2-8\right)}=\dfrac{7t^2+3t}{11t^2-8}\end{matrix}\right.\Rightarrowđpcm\)