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Lời giải:
Đặt \(\frac{a}{2016}=\frac{b}{2018}=\frac{c}{2020}=t\Rightarrow a=2016t; b=2018t; c=2020t\)
Khi đó:
\(\frac{(a-c)^2}{4}=\frac{(2016t-2020t)^2}{4}=\frac{16t^2}{4}=4t^2(1)\)
\((a-b)(b-c)=(2016t-2018t)(2018t-2020t)=(-2t)(-2t)=4t^2(2)\)
Từ \((1);(2)\Rightarrow \frac{(a-c)^2}{4}=(a-b)(b-c)\) (đpcm)
Đặng Quốc Huy:
\(\frac{(2016t-2020t)^2}{4}=\frac{(-4t)^2}{4}=\frac{(-4)^2.t^2}{4}=\frac{16t^2}{4}=4t^2\)
Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\Rightarrow a=2014k;b=2015k;c=2016k\)
=>\(4\left(a-b\right)\left(b-c\right)=4\left(2014k-2015k\right)\left(2015k-2016k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)
\(\left(c-a\right)^2=\left(2016k-2014k\right)^2=\left(2k\right)^2=4k^2\)
=>đpcm
\(\dfrac{a}{2016}=\dfrac{b}{2017}=\dfrac{c}{2018}=\dfrac{a-c}{2016-2018}=\dfrac{a-b}{2016-2017}=\dfrac{b-c}{2017-2018}\)
\(\rightarrow\dfrac{a-c}{-2}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}\)
\(\rightarrow a-c=2\cdot\left(a-b\right)=2\cdot\left(b-c\right)\)
\(\rightarrow\left(a-c\right)^3=\left[2\cdot\left(a-b\right)\right]^2\cdot2\cdot\left(b-c\right)\)
\(\Rightarrow\left(a-c\right)^3=8\cdot\left(a-b\right)^2\cdot\left(b-c\right)\)
Đặt : \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)
\(\Rightarrow\frac{a}{2014}=k\Rightarrow a=2014k\)
\(\Rightarrow\frac{b}{2015}=k\Rightarrow b=2015k\)
\(\Rightarrow\frac{c}{2016}=k\Rightarrow c=2016k\)
Ta có : \(4\left(a-b\right)\left(b-c\right)=4\left(2014k-2015k\right)\left(2015k-2016k\right)\)
\(=4k\left(2014-2015\right).k\left(2015-2016\right)=4k.\left(-1\right).k.\left(-1\right)=4.k^2\)( 1 )
\(\Rightarrow\left(c-a\right)^2=\left(2016k-2014k\right)\left(2016k-2014k\right)=\left[\left(2016k-2014k\right)^2\right]=\left[k\left(2016-2014\right)\right]=\left(k^2\right)^2=k^{2.4}\)( 2 )
Từ \(\left(1\right)\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
1) ADTCDTSBN
có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-7}=\frac{x-y-z}{3-5+7}=\frac{20}{5}=4.\)
=> ...
