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Từ giả thuyết của đề \(\Rightarrow\frac{x}{\frac{1}{a}}=\frac{y}{\frac{1}{b}}=\frac{z}{\frac{1}{c}}=k\)
\(\Rightarrow x=\frac{k}{a};y=\frac{k}{b};z=\frac{k}{c}\)
Mà xyz = \(\frac{8}{abc}=\Rightarrow\frac{k^3}{abc}=\frac{8}{abc}=k^3=8\Rightarrow k=2\)
Vậy \(x=\frac{2}{a};y=\frac{2}{b};z=\frac{2}{c}\)
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ta co ax=by=cz
suy ra x/(1/a)=y/(1/b)=z/(1/c)=k
suy ra x=k/a ; y=k/b ; z=k/c
ma xyz=8/abc suy ra k/a*k/b*k/c=k^3=8
suy ra k=2
suy ra x=2/a;y=2/b;z=2/c
Theo đề ta có: xyz= 8.abc= xyz.abc= ax. by. cz= 8
hay ax.ax.ax= 8
=> (ax)3= 23
=> ax= 2
Với ax= 2=> x= \(\frac{2}{a}\)
by= 2=> y= \(\frac{2}{b}\)
cz= 2=> z=\(\frac{2}{c}\)
Vậy x, y, z= \(\frac{2}{a},\frac{2}{b},\frac{2}{c}.\)
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
ta có: \(ax=by=cz\Rightarrow x:\frac{1}{a}=y:\frac{1}{b}=z:\frac{1}{c}=\frac{x}{\frac{1}{a}}=\frac{y}{\frac{1}{b}}=\frac{z}{\frac{1}{c}}=k.\)
\(\Rightarrow\hept{\begin{cases}x=\frac{k}{a}\\y=\frac{k}{b}\\z=\frac{k}{c}\end{cases}}\)
mà xyz = 8/abc \(\Rightarrow\frac{k}{a}\cdot\frac{k}{b}\cdot\frac{k}{c}=\frac{k^3}{abc}=\frac{8}{abc}\Rightarrow k^3=8=2^3\Rightarrow k=2\)
=> x = 2/a; y = 2/b; z = 2/c