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Find value of Expression

Albert

Well-known member
Jan 25, 2013
1,225
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

Well-known member
Nov 29, 2013
172
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.


[tex]\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=1[/tex]

1º)

[tex]\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0[/tex]

[tex]myz+nxz+pxy=0[/tex]

[tex](mnp)(myz+nxz+pxy)
=0[/tex]

[tex]mnpmyz+mnpnxz+mnppxy=0[/tex](*)

2º)

[tex]\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1[/tex]

[tex]xnp+myp+mnz=mnp[/tex]

[tex](xnp+myp+mnz)^2=m^2n^2p^2[/tex]

[tex]x^2n^2p^2+m^2y^2p^2+m^2n^2z^2+[/tex]

[tex]+2xnpmyp+2xnpmnz+2mypmnz=m^2n^2p^2[/tex]

For (*):

[tex]2xnpmyp+2xnpmnz+2mypmnz=0[/tex]

Then:

[tex]x^2n^2p^2+m^2y^2p^2+m^2n^2z^2=m^2n^2p^2[/tex]

[tex]\dfrac{x^2n^2p^2+m^2y^2p^2+m^2n^2z^2}{m^2n^2p^2}=1[/tex]

[tex]\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=1[/tex]


Regards.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Albert!

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

. . . . . . . [tex]\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0[/tex]

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

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

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

[tex]\text{And we want: }\:a^2+b^2+c^2.[/tex]


[tex]\text{From [2]: }\:\frac{ab+bc+ac}{abc} \:=\:0 \quad\Rightarrow\quad ab + bc + ac \:=\:0[/tex]

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

. . [tex]a^2+2ab+2ac+b^2+2bc+c^2 \:=\:1[/tex]

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

[tex]\text{Therefore: }\:a^2+b^2+c^2 \:=\:1[/tex]
 
Last edited by a moderator:

Albert

Well-known member
Jan 25, 2013
1,225
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)?
$\dfrac {m^2}{x^2}+\dfrac{n^2}{y^2} +\dfrac {p^2}{z^2}=?$