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)

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

$$myz+nxz+pxy=0$$

$$(mnp)(myz+nxz+pxy) =0$$

$$mnpmyz+mnpnxz+mnppxy=0$$(*)

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

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

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

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

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

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

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

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

 

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

 

[CellCoree]

To find the value of this expression, we can use the given equations to solve for the values of x, y, and z in terms of m, n, and p. From the first equation, we can rearrange to get x = m(1 - y/n - z/p). Substituting this into the second equation, we get m(1 - y/n - z/p)^2/m + n/y + p/z = 0. Simplifying, we get 1 - y/n - z/p = 0. This means that y/n + z/p = 1. Substituting this into the first equation, we get x/m + 1 = 1, which means x = 0. Similarly, we can solve for y and z to get y = 0 and z = 0.

Substituting these values into the expression we are trying to find, we get (0)^2/(m^2) + (0)^2/(n^2) + (0)^2/(p^2) = 0. Therefore, the value of the expression is 0.