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Theo tc dãy tỉ số bằng nhau
\(\frac{a-6b}{3c}=\frac{2b-9c}{a}=\frac{3c-3a}{2b}=\frac{a+2b+3c-6b-9c-3a}{3c+a+2b}\)
\(=\frac{a+2b+3a-3\left(2b+3c+a\right)}{3c+a+2b}=\frac{-2.72}{72}=-2\)
\(\Rightarrow a-6b=-6c;3c-3a=-4b\Leftrightarrow3a-4b=3c\)
ta có hệ \(\hept{\begin{cases}a-6b=-6c\\3a-4b=3c\end{cases}\Leftrightarrow\hept{\begin{cases}3a-18b=-18c\\3a-4b=3c\end{cases}}\Leftrightarrow\hept{\begin{cases}-14b=-21c\left(1\right)\\a=-6c+6b\left(2\right)\end{cases}}}\)
Theo giả thiết \(a+2b+3c=72\Rightarrow a=-2b-3c-72\)
\(\Rightarrow-2b-3c-72=-6c+6b\Leftrightarrow8b-3c+72=0\Leftrightarrow8b-3c=-72\)
(1) => \(\frac{b}{-21}=\frac{c}{-14}\)Theo tc dãy tỉ số bằng nhau
\(\frac{b}{-21}=\frac{c}{-14}=\frac{8b-3c}{8\left(-21\right)-3\left(-14\right)}=-\frac{72}{-126}=\frac{4}{7}\Rightarrow b=-12;c=-8\)
Thay vào (2) vậy \(a=-6c+6b=-6\left(-8\right)+6\left(-12\right)=48-72=-24\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=k\Rightarrow a=bk;b=ck;c=dk;d=ek\)
\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)
\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)
Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)
Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
Giải:
Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,b=ck,c=dk\)
Ta có:
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{bk+ck-dk}{b+c-d}\right)^3=\left[\frac{k\left(b+c-d\right)}{b+c-d}\right]^3=k^3\) (1)
\(\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^2=\left(\frac{2bk+3ck-4dk}{2b+3c-4d}\right)^3=\left[\frac{k\left(2b+3c-4d\right)}{2b+3c-4d}\right]^3=k^3\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^3\) ( đpcm )
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{2a+3c}{3a+4c}=\dfrac{2bk+3dk}{3bk+4dk}=\dfrac{2b+3d}{3b+4d}\)