What is the value of the expression given these equations?

[Albert1]

given :

$\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1$

$\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0$

find:

$\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=?$

 

[mente oscura]

Re: find value

given :

$\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1$

$\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0$

find:

$\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=?$

Hello.

Spoiler

\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=1

1º)

\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0

myz+nxz+pxy=0

(mnp)(myz+nxz+pxy)<br /> =0

mnpmyz+mnpnxz+mnppxy=0(*)

2º)

\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1

xnp+myp+mnz=mnp

(xnp+myp+mnz)^2=m^2n^2p^2

x^2n^2p^2+m^2y^2p^2+m^2n^2z^2+

+2xnpmyp+2xnpmnz+2mypmnz=m^2n^2p^2

For (*):

2xnpmyp+2xnpmnz+2mypmnz=0

Then:

x^2n^2p^2+m^2y^2p^2+m^2n^2z^2=m^2n^2p^2

\dfrac{x^2n^2p^2+m^2y^2p^2+m^2n^2z^2}{m^2n^2p^2}=1

\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=1

Regards.

 

[soroban]

Hello, Albert!

\text{Given: }\:\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1

. . . . . . . \dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0

\text{Find: }\:\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}

Spoiler

\text{Let: }\:a\,=\,\frac{x}{m},\;\;b \,=\, \frac{y}{n},\;\;c \,=\, \frac{z}{p}

\text{We have: }\:\begin{Bmatrix}a+b+c &amp;=&amp; 1 &amp; [1] \\ \frac{1}{a}+\frac{1}{b} + \frac{1}{c} &amp;=&amp; 0 &amp; [2] \end{Bmatrix}

\text{And we want: }\:a^2+b^2+c^2.\text{From [2]: }\:\frac{ab+bc+ac}{abc} \:=\:0 \quad\Rightarrow\quad ab + bc + ac \:=\:0

\text{Square [1]: }\: (a+b+c)^2 \:=\:1 ^2

. . a^2+2ab+2ac+b^2+2bc+c^2 \:=\:1

. . a^2+b^2+c^2 + 2\underbrace{(ab + bc + ac)}_{\text{This is }0} \:=\:1

\text{Therefore: }\:a^2+b^2+c^2 \:=\:1

 

[Albert1]

given :

$\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1$

$\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0$

find:

$\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=?$

now the following value can be found (is it a fixed number)?
$\dfrac {m^2}{x^2}+\dfrac{n^2}{y^2} +\dfrac {p^2}{z^2}=?$