Question

Bài 1

\(\frac{x^3}{8}=\frac{^{y^3}}{64}=\frac{z^3}{2016}\) và \(x^2+y^2+z^2\)=14

tìm x,y,z

Bài 2

Cho \(a^2\)=b.c

CM \(\frac{a+b}{a-c}=\frac{c+a}{c-a}\)

Answer 1

Mk sửa lại đề nhá : \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\)

Ta có :  \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}=\left(\frac{x}{2}\right)^3=\left(\frac{y}{4}\right)^3=\left(\frac{z}{6}\right)^3\)

\(\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)

Đặt t = \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)

Khi đó \(t^2=\frac{x^2}{2^2}=\frac{y^2}{4^2}=\frac{z^2}{6^2}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

Suy ra : t = \(\frac{1}{2};-\frac{1}{2}\)

+ t = \(\frac{1}{2}\) thì x = \(\frac{1}{2}\).2 = 1

                      y = \(\frac{1}{2}\).4 = 2

                      z = \(\frac{1}{2}\).6 = 3

+ t = \(-\frac{1}{2}\) thì x = \(-\frac{1}{2}\). 2 = -1

                         y ​\(=-\frac{1}{2}.4=-2\)

                        z \(=-\frac{1}{2}.6=-3\)