Question

[nguyen ngoc thanh huong]

tìm 3 số hữu tỉ x,y,z biết xy=\(\frac{1}{3}\), yz=\(\frac{-2}{5}\)và zx=\(\frac{-3}{10}\)

Answer 1

[Lê Song Thanh Nhã]

Ta có: xy.yz.zx = \(\frac{1}{3}\times\frac{-2}{5}\times\frac{-3}{10}=\frac{1}{25}\)=> \(\left(xyz\right)^2=\frac{1}{25}\)

Mà \(\frac{1}{25}=\left(\frac{1}{5}\right)^2=\left(-\frac{1}{5}\right)^2\)

Nếu \(\left(xyz\right)^2=\left(\frac{1}{5}\right)^2\Rightarrow xyz=\frac{1}{5}\)

=> \(x=\frac{1}{5}:yz=\frac{1}{5}:\left(-\frac{2}{5}\right)=-\frac{1}{2}\)

=> \(y=\frac{1}{5}:xz=\frac{1}{5}:\left(-\frac{3}{10}\right)=-\frac{2}{3}\)

=> \(z=\frac{1}{5}:xy=\frac{1}{5}:\frac{1}{3}=\frac{3}{5}\)

Nếu \(\left(xyz\right)^2=\left(-\frac{1}{5}\right)^2\Rightarrow xyz=-\frac{1}{5}\)

(Tương tự trên nha ^^ )

Answer 2

[Lê Chí Cường]

=>\(xy.yz.zx=\frac{1}{3}.\frac{-2}{5}.\frac{-3}{10}=\frac{6}{150}=\frac{1}{25}\)

=>\(x^2.y^2.z^2=\frac{1^2}{5^2}\)

=>\(\left(x.y.z\right)^2=\left(\frac{1}{5}\right)^2\)

=>\(x.y.z=\frac{1}{5}\)

=>\(x=\frac{1}{5}:\frac{-2}{5}=\frac{-1}{2}\)

=>\(y=\frac{1}{5}:\frac{-3}{10}=\frac{-2}{3}\)

=>\(z=\frac{1}{5}:\frac{1}{3}=\frac{3}{5}\)