Ensino Fundamental ⇒ Algebra Tópico resolvido

[botelho]

Tem-se: [tex3]\frac{a}{b+c+d} + \frac{b}{a+c+d} + \frac{c}{a+b+d} + \frac{d}{a+b+c}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{a}{b+c+d} + \frac{b}{a+c+d} + \frac{c}{a+b+d} + \frac{d}{a+b+c}[/tex3]

=1.Calcule [tex3]\frac{a²}{b+c+d} + \frac{b²}{a+c+d} + \frac{c²}{a+b+d} + \frac{d²}{a+b+c}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{a²}{b+c+d} + \frac{b²}{a+c+d} + \frac{c²}{a+b+d} + \frac{d²}{a+b+c}[/tex3]

.
a)0
b)1
c)2
d)4
e)[tex3]\frac{1}{4}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{4}[/tex3]

Resposta

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a

[deOliveira]

[tex3]\frac a{b+c+d}+\frac b{a+c+d}+\frac c{a+b+d}+\frac d{a+b+c}=1[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac a{b+c+d}+\frac b{a+c+d}+\frac c{a+b+d}+\frac d{a+b+c}=1[/tex3]

Vamos multiplicar essa igualdade por [tex3]a,b,c,d[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]a,b,c,d[/tex3]

separadamente obtendo as quatro seguintes igualdades:

[tex3](I):\ \frac {a^2}{b+c+d}+\frac {ab}{a+c+d}+\frac {ac}{a+b+d}+\frac {ad}{a+b+c}=a\\(II): \frac {ab}{b+c+d}+\frac {b^2}{a+c+d}+\frac {bc}{a+b+d}+\frac {bd}{a+b+c}=b\\(III):\ \frac {ac}{b+c+d}+\frac {bc}{a+c+d}+\frac {c^2}{a+b+d}+\frac {cd}{a+b+c}=c\\(IV):\ \frac {ad}{b+c+d}+\frac {bd}{a+c+d}+\frac {cd}{a+b+d}+\frac {d^2}{a+b+c}=d[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3](I):\ \frac {a^2}{b+c+d}+\frac {ab}{a+c+d}+\frac {ac}{a+b+d}+\frac {ad}{a+b+c}=a\\(II): \frac {ab}{b+c+d}+\frac {b^2}{a+c+d}+\frac {bc}{a+b+d}+\frac {bd}{a+b+c}=b\\(III):\ \frac {ac}{b+c+d}+\frac {bc}{a+c+d}+\frac {c^2}{a+b+d}+\frac {cd}{a+b+c}=c\\(IV):\ \frac {ad}{b+c+d}+\frac {bd}{a+c+d}+\frac {cd}{a+b+d}+\frac {d^2}{a+b+c}=d[/tex3]

Fazendo [tex3](I)+(II)+(III)+(IV)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3](I)+(II)+(III)+(IV)[/tex3]

temos:

[tex3]a+b+c+d=\frac{ab+bc+bd}{a+c+d}+\frac{ac+bc+cd}{a+b+d}+\frac{ad+bd+cd}{a+b+c}+\frac{ab+ac+ad}{b+c+d}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
a+b+c+d=\frac{b{(a+c+d)}}{a+c+d}+\frac{c{(a+b+d)}}{a+b+d}+\frac{d{(a+b+c)}}{a+b+c}+\frac{a{(b+c+d)}}{b+c+d}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
a+b+c+d=\frac{b\cancel{(a+c+d)}}{\cancel{a+c+d}}+\frac{c\cancel{(a+b+d)}}{\cancel{a+b+d}}+\frac{d\cancel{(a+b+c)}}{\cancel{a+b+c}}+\frac{a\cancel{(b+c+d)}}{\cancel{b+c+d}}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
a+b+c+d=a+b+c+d+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
\cancel{a+b+c+d}=\cancel{a+b+c+d}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\\therefore\boxed{\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}=0}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]a+b+c+d=\frac{ab+bc+bd}{a+c+d}+\frac{ac+bc+cd}{a+b+d}+\frac{ad+bd+cd}{a+b+c}+\frac{ab+ac+ad}{b+c+d}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
a+b+c+d=\frac{b{(a+c+d)}}{a+c+d}+\frac{c{(a+b+d)}}{a+b+d}+\frac{d{(a+b+c)}}{a+b+c}+\frac{a{(b+c+d)}}{b+c+d}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
a+b+c+d=\frac{b\cancel{(a+c+d)}}{\cancel{a+c+d}}+\frac{c\cancel{(a+b+d)}}{\cancel{a+b+d}}+\frac{d\cancel{(a+b+c)}}{\cancel{a+b+c}}+\frac{a\cancel{(b+c+d)}}{\cancel{b+c+d}}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
a+b+c+d=a+b+c+d+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\
\cancel{a+b+c+d}=\cancel{a+b+c+d}+\left[\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}\right]\\\therefore\boxed{\frac {a^2}{b+c+d}+\frac {b^2}{a+c+d}+\frac {c^2}{a+b+d}+\frac {d^2}{a+b+c}=0}[/tex3]

Espero ter ajudado .