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\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
Theo tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)
Ta lại có: \(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)
Từ (1) và (2) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
Ta có :
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{b^2}{c}\\d=\dfrac{c^2}{b}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{b^2}{c}:\dfrac{c^2}{b}\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{b^2}{c}.\dfrac{b}{c^2}\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{b^3}{c^3}=\dfrac{8b^3}{8c^3}=\dfrac{a^3}{b^3}=\dfrac{125c^3}{125d^3}\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{a^3+8b^3+125c^3}{b^3+8c^3+125d^3}\left(dpcm\right)\)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\) (1)
Ta lại có : \(\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\) (2)
Từ (1) ; (2) => \(\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\) (ĐPCM)
Cách 1:
Ta xét tích a(c-d) và c(a-b)
Ta có: a(c-d)=ac-ad (1)
c(a-b)=ac-bc(2)
Ta lại có \(\dfrac{a}{c}=\dfrac{c}{d}\)=>ad=bc (3)
Từ (1), (2), (3) ta có a(c-d)=c(a-d). Do đó \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cách 2:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k thì a=bk, c=dk.
Xét \(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
Xét \(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cách 3: Ta có
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Aps dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a-b}{c-d}\)
=>\(\dfrac{a}{c}=\dfrac{a-b}{c-d}=>\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
hay \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c};c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{c^3+b^3+d^3}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;b=ck;c=dk\\ \Rightarrow a=bk=ck^2=dk^3\\ \Rightarrow\dfrac{a}{d}=k^3\\ \text{Mà }\dfrac{a}{b}=k\Rightarrow\dfrac{a^3}{b^3}=k^3\\ \Rightarrow\dfrac{a}{d}=\dfrac{a^3}{b^3}\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)