Ensino Médio ⇒ Produtos Notáveis e Fatoração Tópico resolvido

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Se a = xy/x + y, b = yz/y + z, c = zx/z + x, em que a, b, c, são diferentes de 0, então x é igual a:
a) abc/ab + bc + ca
b) 2abc/ab + bc + ca
c) 2abc/ab - bc - ca
d) 2abc/-ab + bc + ca
e) 2abc/ab + bc - ca
Resposta e)

[petras]

[tex3]\\ c=\frac{z.x}{(z+x)}\rightarrow c.x+c.z = x.z\rightarrow cx = (x-c).z\rightarrow z=\frac{c.x}{(x-c)} \\\
b = \frac{y.z}{(y+z)}b.y + b.z = y.z \rightarrow b.y=(y-b).z \rightarrow z=\frac{b.y}{(y-b)}\\\
Portanto:\frac{c.x}{(x-c)}= \frac{b.y}{(y-b)}\rightarrow (y-b).(c.x)=(x-c).b.y\rightarrow y.c.x-b.c.x=b.x.y-b.c.y\rightarrow \\\ y.c.x-b.x.y+b.c.y=b.c.x\rightarrow y(c.x-b.x+b.c) = b.c.x \rightarrow y=\frac{b.c.x}{c.x-b.x+b.c}\\\
Mas\ a = \frac{x.y}{(x+y)}\rightarrow a = \frac{x.(\frac{bcx}{cx-bx+bc})}{(x+\frac{bcx}{cx-bx+bc})} \rightarrow a = \frac{x.(\frac{bcx}{cx-bx+bc})}{\frac{x(cx-bx+bc)+bcx}{cx-bx+bc}} \rightarrow a=\frac{x.(bcx)}{x.({cx-bx+bc})+bcx}\rightarrow a=\frac{(bcx)}{x.({cx-bx+bc}+bc)}\rightarrow\\
a=\frac{bcx}{cx-bx+2bc}\rightarrow acx-abx+2abc=bcx\rightarrow acx-abx-bcx=-2abc\rightarrow x(ac-ab-bc)=-2abc\\
x=\frac{-2abc}{(ac-ab-bc)}.(\frac{-1}{-1})\rightarrow \ \boxed{\mathsf{ x=\frac{2abc}{(ab+bc-ca)} }}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\\ c=\frac{z.x}{(z+x)}\rightarrow c.x+c.z = x.z\rightarrow cx = (x-c).z\rightarrow z=\frac{c.x}{(x-c)} \\\
b = \frac{y.z}{(y+z)}b.y + b.z = y.z \rightarrow b.y=(y-b).z \rightarrow z=\frac{b.y}{(y-b)}\\\
Portanto:\frac{c.x}{(x-c)}= \frac{b.y}{(y-b)}\rightarrow (y-b).(c.x)=(x-c).b.y\rightarrow y.c.x-b.c.x=b.x.y-b.c.y\rightarrow \\\ y.c.x-b.x.y+b.c.y=b.c.x\rightarrow y(c.x-b.x+b.c) = b.c.x \rightarrow y=\frac{b.c.x}{c.x-b.x+b.c}\\\
Mas\ a = \frac{x.y}{(x+y)}\rightarrow a = \frac{x.(\frac{bcx}{cx-bx+bc})}{(x+\frac{bcx}{cx-bx+bc})} \rightarrow a = \frac{x.(\frac{bcx}{cx-bx+bc})}{\frac{x(cx-bx+bc)+bcx}{cx-bx+bc}} \rightarrow a=\frac{x.(bcx)}{x.({cx-bx+bc})+bcx}\rightarrow a=\frac{(bcx)}{x.({cx-bx+bc}+bc)}\rightarrow\\
a=\frac{bcx}{cx-bx+2bc}\rightarrow acx-abx+2abc=bcx\rightarrow acx-abx-bcx=-2abc\rightarrow x(ac-ab-bc)=-2abc\\
x=\frac{-2abc}{(ac-ab-bc)}.(\frac{-1}{-1})\rightarrow \ \boxed{\mathsf{ x=\frac{2abc}{(ab+bc-ca)} }}[/tex3]