Đặt a/2019=b/2021=c/2023=k

=>a=2019k; b=2021k; c=2023k

(a-c)^2/4=(2023k-2019k)^2/4=(4k)^2/4=4k^2

(a-b)(b-c)=(2019k-2021k)(2021k-2023k)=4k^2

=>(a-c)^2/4=(a-b)(b-c)

Ta có:  \(\frac{a}{n+2}=\frac{b}{n+5}=\frac{c}{n+8}\)

  \(\Rightarrow\frac{a}{n+2}=\frac{b}{n+5}=\frac{c}{n+8}=\frac{a-c}{-6}=\frac{b-c}{-3}=\frac{a-b}{-3}\)

  Đặt \(\frac{a-c}{-6}=\frac{b-c}{-3}=\frac{a-b}{-3}=k\)

         \(\Rightarrow a-c=-6k\) ; \(b-c=-3k\) ; \(a-b=-3k\)

Thay vào 2 biểu thức, ta có:

 \(\left(a-c\right)^2=\left(-6k\right)^2=36k^2\) (1)

 \(4\left(a-b\right)\left(b-c\right)=4.\left(-3k\right).\left(-3k\right)=4.\left(-3k\right)^2=4.9k^2=36k^2\) (2)

 Từ (1) và (2), suy ra \(\left(a-c\right)^2=4\left(a-b\right)\left(b-c\right)\)

Lời giải:

Ta có:

$\frac{a}{b}=\frac{c}{d}=\frac{4c}{4d}=\frac{a+4c}{b+4d}$ (theo TCDTSBN)

$\frac{a}{b}=\frac{c}{d}=\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a-3c}{2b-3d}$ (theo TCDTSBN)

$\Rightarrow \frac{a+4c}{b+4d}=\frac{2a-3c}{2b-3d}$

$\Rightarrow (a+4c)(2b-3d)=(2a-3c)(b+4d)$ (đpcm)

đặt a/2003=b/2005=c/2007=t

=>a=2003t;b=2005t;c=2007t

ta có:\(VT=\frac{\left(a-c\right)^2}{4}=\frac{\left(2003t-2007t\right)^2}{4}=\frac{\left(-4t\right)^2}{4}=\frac{\left(-4\right)^2.t^2}{4}=\frac{16.t^2}{4}=\frac{4.4.t^2}{4}=4t^2\) (1)

\(VP=\left(a-b\right)\left(b-c\right)=\left(2003t-2005t\right)\left(2005t-2007t\right)=\left(-2\right).t.\left(-2\right).t=\left[\left(-2\right).\left(-2\right)\right].t^2=4t^2\left(2\right)\)

từ (1);(2) ta có VT=VP=>đpcm