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\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=t\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)
Ta có đpcm
Ta có :
\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(3\right)\)
Lại có :
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(4\right)\)
Từ \(\left(3\right)+\left(4\right)\Leftrightarrowđpcm\)
Ta co :\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)=>\(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
=> \(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (1)
Mặt khác:\(\dfrac{a^3}{b^3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\) (2)
Tu (1) va (2)
=> \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\) (dpcm)
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{ac}{c^2}\)=\(\dfrac{bd}{d^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{d^2}{c^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)= \(\dfrac{2c^2-ac}{2c^2-bd}\)
=> \(\dfrac{a}{b}\)=\(\dfrac{2c^2-ac}{2c^2-bd}\)=>\(\dfrac{a^2}{b^2}\)=\(\dfrac{2c^2-ac}{2d^2-bd}\)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)= \(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)=\(\dfrac{ma+nb}{mc+nd}\)=\(\dfrac{ma-nb}{mc-nd}\)
=> \(\dfrac{ma+nb}{ma-nb}\)=\(\dfrac{mc+nd}{mc-nd}\)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^3}{c^3}\)=\(\dfrac{b^3}{d^3}\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^3\)(2)
Từ (1) và (2) suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^3\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)
Giải:
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\left(đpcm\right)\)
Vậy...
Giải:
Vì:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}_{\left(2\right)}.\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}.\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{abc}{bcd}=\dfrac{a}{d}.\)
\(\Rightarrowđpcm.\)
Ta có: \(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)
\(\dfrac{a^3}{b^3}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)
Vậy ............................
\(\rightarrowđpcm\)