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Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)
Ta có: \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{ak\cdot bk\cdot ck\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\cdot\left(ak+bk\right)\cdot\left(bk+ck\right)\cdot\left(ck+ak\right)}\)
\(=\frac{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)
Vậy: H=1
đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)
theo giả thiết ta có \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
thay \(H=\frac{ak.bk.ck\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(ak+bk\right)\left(bk+ck\right)\left(ck+ak\right)}\)
\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left[k\left(a+b\right)\right]\left[k\left(b+c\right)\right]\left[k\left(c+a\right)\right]}\)
\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc.k\left(a+b\right).k\left(b+c\right).k\left(c+a\right)}\)
\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)
Vậy H = 1
\(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\Leftrightarrow\frac{y+z}{\frac{1}{a}}=\frac{z+x}{\frac{1}{b}}=\frac{x+y}{\frac{1}{c}}=\)
\(=\frac{y+z-\left(z+x\right)}{\frac{1}{a}-\frac{1}{b}}=\frac{z+x-\left(x+y\right)}{\frac{1}{b}-\frac{1}{c}}=\frac{x+y-\left(y+z\right)}{\frac{1}{c}-\frac{1}{a}}=\frac{y-x}{\frac{b-a}{ab}}=\frac{z-y}{\frac{c-b}{bc}}=\frac{x-z}{\frac{a-c}{ac}}\)
Chia các vế của 3 tỷ lệ thức cuối cho abc ta có:
\(\frac{y-x}{\frac{b-a}{ab}\cdot abc}=\frac{z-y}{\frac{c-b}{bc}\cdot abc}=\frac{x-z}{\frac{a-c}{ac}\cdot abc}=\frac{y-x}{c\left(b-a\right)}=\frac{z-y}{a\left(c-b\right)}=\frac{x-z}{b\left(a-c\right)}\)
Hay: \(\frac{x-y}{c\left(a-b\right)}=\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}\)đpcm
Ta có:
\(\dfrac{a.\left(x+z\right)}{abc}=\dfrac{b.\left(z+x\right)}{abc}=\dfrac{c.\left(x+y\right)}{abc}\)
\(\Rightarrow\dfrac{y+z}{bc}=\dfrac{x+z}{ac}=\dfrac{x+y}{ab}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z}{bc}=\dfrac{x+z}{ac}=\dfrac{x+y}{ab}=\dfrac{z+x-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c.\left(a-b\right)}\left(1\right)\)
\(\dfrac{y+z}{bc}=\dfrac{x+z}{ac}=\dfrac{x+y}{ab}=\dfrac{y+z-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b.\left(c-a\right)}\left(2\right)\)
\(\dfrac{y+z}{bc}=\dfrac{x+z}{ac}=\dfrac{x+y}{ab}=\dfrac{x+y-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a.\left(b-c\right)}\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\) suy ra:
\(\dfrac{y-z}{a.\left(b-c\right)}=\dfrac{z-x}{b.\left(c-a\right)}=\dfrac{x-y}{c.\left(a-b\right)}\)
\(\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(\Leftrightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=c-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b-c=0\\c-a=0\\a-b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=c\\c=a\\a=b\end{cases}}\)
\(\Leftrightarrow a=b=c\)( đpcm )
\(\Rightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=a-c\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}b-c=0\\c-a=0\\a-b=0\end{cases}\Rightarrow\hept{\begin{cases}b=c\\c=a\\a=b\end{cases}\Rightarrow}a=b=c\left(dpcm\right)}\)
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,