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Đặt \(\frac{a}{2008}=\frac{b}{2009}=\frac{c}{2010}=k\)
suy ra: \(a=2008k;\) \(b=2009k;\)\(c=2010k\)
Khi đó ta có: \(4\left(a-b\right)\left(b-c\right)\)
\(=4\left(2008k-2009k\right)\left(2009k-2010k\right)\)
\(=4k^2\)
\(\left(c-a\right)^2=\left(2010k-2008k\right)^2=4k^2\)
suy ra: \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\) (đpcm)
p/s: tham khảo,
đặt a/2008=b/2009=c/2010=k=>a=2008k;b=2009k;c=2010k
thay vào biểu thức:
\(\left(a-c\right)^3:\left[\left(a-b\right)^2.\left(b-c\right)\right]=\left(2008k-2010k\right)^3:\left[\left(2008k-2009k\right)^2.\left(2009k-2010k\right)\right]\)
\(=\left(-2k\right)^3:\left[\left(-1k\right)^2.\left(-1k\right)^2\right]=\left(-2\right)^3.k^3:\left[\left(-1\right)^2.k^2.\left(-1\right)^2.k^2\right]=8.k^3:1.k^4=8.k^3:k^4=8.k^3:k^3.k=8k\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)
\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)
\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)
=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)
=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Ta có: \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=\frac{a-b}{2009-2010}=\frac{b-c}{2010-2011}=\frac{c-a}{2011-2009}.\)
\(\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)
\(\Rightarrow\frac{a-b}{-1}.\frac{b-c}{-1}=\left(\frac{c-a}{2}\right)^2\)
\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{2^2}\)
\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{4}.\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2.1\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)-\left(c-a\right)^2=0.\)
Hay \(M=0.\)
Vậy \(M=0.\)
Chúc bạn học tốt!
Bài 2 )
\(a\left(y+z\right)=b\left(x+z\right)=c\left(x+y\right)\)
\(\Leftrightarrow\frac{a\left(y+z\right)}{abc}=\frac{b\left(x+z\right)}{abc}=\frac{c\left(x+y\right)}{abc}\)
\(\Leftrightarrow\frac{y+z}{bc}=\frac{x+z}{ac}=\frac{x+y}{ab}\)
\(\Leftrightarrow\frac{bc}{y+z}=\frac{ac}{x+z}=\frac{ab}{x+y}\)
Đặt \(\frac{bc}{y+z}=\frac{ac}{x+z}=\frac{ab}{x+y}=k\)
\(\Rightarrow\left\{\begin{matrix}bc=k\left(y+z\right)=ky+kz\\ac=k\left(x+z\right)=kx+kz\\ab=k\left(x+y\right)=kx+ky\end{matrix}\right.\) (1)
Gỉa sử điều cần chứng minh là đúng ta có
\(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
\(\Leftrightarrow\frac{y-z}{ab-ac}=\frac{z-x}{bc-ab}=\frac{x-y}{ac-bc}\)
Thế (1) vào biểu thức
\(\frac{y-z}{kx+ky-\left(kx+kz\right)}=\frac{z-x}{ky+kz-\left(kx+ky\right)}=\frac{x-y}{kx+kz-\left(ky+kz\right)}\)
\(\Leftrightarrow\frac{y-z}{ky-kz}=\frac{z-x}{kz-kx}=\frac{x-y}{kx-ky}\)
\(\Leftrightarrow\frac{y-z}{k\left(y-z\right)}=\frac{z-x}{k\left(z-x\right)}=\frac{x-y}{k\left(x-y\right)}\)
\(\Leftrightarrow\frac{1}{k}=\frac{1}{k}=\frac{1}{k}\) ( điều này luôn luôn đúng )
\(\Rightarrow\) ĐPCM
Ta có : \(\frac{a+b}{2008}=\frac{b+c}{2009}=\frac{c+a}{2010}=\frac{a+b-\left(b+c\right)}{2008-2009}=\frac{b+c-\left(c+a\right)}{2009-2010}=\frac{c+a-\left(a+b\right)}{2010-2008}=\frac{a-c}{-1}=\frac{b-a}{-1}=\frac{c-b}{2}\)
Đặt \(\frac{a-c}{-1}=\frac{b-a}{-1}=\frac{c-b}{2}=k\Rightarrow a-c=-k;b-a=-k;c-b=2k\)
Ta lại có : \(4\left(a-c\right)\left(b-a\right)=\left(c-b\right)^2\)\(\Rightarrow-4k\times\left(-k\right)=\left(2k\right)^2\)\(\Rightarrow4k^2=4k^2\)
Vế trái đúng bằng vế phải \(\Rightarrow\)\(4\left(a-c\right)\left(b-a\right)=\left(c-b\right)^2\)