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ta có x.y.z=8/a.b.c->ax.by.cz=8 hay ax.ax.ax=8 <-> (ax)3=23
--->ax=2-->x=2/a,y=2/b,z=2/c
có gì ko hiểu hỏi anh nhé
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ta co :
xyz=8/abc=>ax.by.cz=8 hay ax.ax.ax=8 <=> (ax)^3=2^3
=>ax=2=>x/a;y=2/b;z=2/c
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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
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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
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ax = by <=> x/b = y/a <=> \(\frac{x}{bc}=\frac{y}{ac}\)
Tương tự by = cz <=> \(\frac{y}{c}=\frac{z}{b}\) <=> \(\frac{y}{ac}=\frac{z}{ab}\)
=> \(\frac{x}{bc}=\frac{y}{ac}=\frac{z}{ab}\)<=> \(\frac{xyz}{a^2.b^2.c^2}=\frac{\frac{8}{abc}}{a^2.b^2.c^2}=\frac{8}{a^3.b^3.c^3}\)<=>\(x.y.z=\frac{2}{a}.\frac{2}{b}.\frac{2}{c}\)=> x = 2/a ,y = 2/b , z = 2/ c
Mình thấy đề bài nó "sao sao" ấy bạn! Kệ cứ giải đại vậy!
Ta có: \(ax=by=cz\Rightarrow\hept{\begin{cases}\frac{x}{b}=\frac{y}{a}\\\frac{y}{c}=\frac{z}{b}\end{cases}}\Leftrightarrow\frac{x}{bc}=\frac{y}{ac}=\frac{z}{ba}\)
Đặt \(\frac{x}{bc}=\frac{y}{ac}=\frac{z}{ba}=k\Rightarrow x=kbc,y=kac,z=kba\)
\(\Rightarrow xyz=kbc.kac.kba=k^3.a^2.b^2.c^2=k^3\left(abc\right)^2=\frac{108}{abc}\)
Ta được: \(k^3\left(abc\right)^2=\frac{108}{abc}\). Nhân cả hai vế với abc để khử mẫu,ta có:
\(k^3\left(abc\right)^{\text{3 }}=108\Leftrightarrow k^3=\frac{108}{\left(abc\right)^3}\Leftrightarrow k=\sqrt[3]{\frac{108}{\left(abc\right)^3}}\)
Suy ra:\(\hept{\begin{cases}x=\sqrt[3]{\frac{108}{\left(abc\right)^3}}.bc\\y=\sqrt[3]{\frac{108}{\left(abc\right)^3}}.ac\\z=\sqrt[3]{\frac{108}{\left(abc\right)^3}}.ba\end{cases}}\)