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Montrer-que-x-an-bm-m-n-




Question Number 197541 by a.lgnaoui last updated on 20/Sep/23
Montrer que       x=((an+bm)/(m+n))
$$\boldsymbol{\mathrm{Montrer}}\:\boldsymbol{\mathrm{que}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{an}}+\boldsymbol{\mathrm{bm}}}{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} \\ $$
Commented by a.lgnaoui last updated on 20/Sep/23
Commented by a.lgnaoui last updated on 21/Sep/23
Commented by a.lgnaoui last updated on 21/Sep/23
Answered by HeferH last updated on 20/Sep/23
 ((x−a)/(b−x)) = (m/n)     xn−an = bm−mx   x(n+m) = an+bm   x = ((an+bm)/(m+n))
$$\:\frac{{x}−{a}}{{b}−{x}}\:=\:\frac{{m}}{{n}}\:\: \\ $$$$\:{xn}−{an}\:=\:{bm}−{mx} \\ $$$$\:{x}\left({n}+{m}\right)\:=\:{an}+{bm} \\ $$$$\:{x}\:=\:\frac{{an}+{bm}}{{m}+{n}}\: \\ $$
Answered by a.lgnaoui last updated on 21/Sep/23
propriete des triangles semblables:  (meme angle(𝛂) et 2 cotes paraleles)  ((x−a)/m)=((b−x)/n)    ⇒  m(b−x)=n(x−a)  (m+n)x=an+bm    x=((an+bm)/(m+n))
$$\boldsymbol{\mathrm{propriete}}\:\boldsymbol{\mathrm{des}}\:\boldsymbol{\mathrm{triangles}}\:\boldsymbol{\mathrm{semblables}}: \\ $$$$\left(\boldsymbol{\mathrm{meme}}\:\boldsymbol{\mathrm{angle}}\left(\boldsymbol{\alpha}\right)\:\mathrm{et}\:\mathrm{2}\:\boldsymbol{\mathrm{cotes}}\:\boldsymbol{\mathrm{paraleles}}\right) \\ $$$$\frac{\mathrm{x}−\mathrm{a}}{\mathrm{m}}=\frac{\mathrm{b}−\mathrm{x}}{\mathrm{n}}\:\:\:\:\Rightarrow\:\:\mathrm{m}\left(\mathrm{b}−\mathrm{x}\right)=\mathrm{n}\left(\mathrm{x}−\mathrm{a}\right) \\ $$$$\left(\mathrm{m}+\mathrm{n}\right)\mathrm{x}=\mathrm{an}+\mathrm{bm}\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{an}}+\boldsymbol{\mathrm{bm}}}{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} \\ $$

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