Câu hỏi của Nguyễn Minh Hoàng

Question

Tìm x, y, z biết:

ax = by = cz và $xyz=\frac{8}{abc}\left(a,b,c\ne0\right)$

I don't need you · Accepted Answer

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

Câu trả lời:

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