Ensino Fundamental ⇒ Produtos Notáveis Tópico resolvido

[botelho]

Se a+b+c=12 e a³+b³+c³=1440, calcule o valor de [tex3]\frac{1}{12a+bc} + \frac{1}{12b+ac} + \frac{1}{12c+ab}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{12a+bc} + \frac{1}{12b+ac} + \frac{1}{12c+ab}[/tex3]

.
a)[tex3]\frac{1}{4}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

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

b)[tex3]\frac{1}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{2}[/tex3]

c)[tex3]\frac{1}{3}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{3}[/tex3]

d)2
e)1

Resposta

---

a

[petras]

Resposta do colega Jonathan:

Expandindo [tex3]\mathsf{\frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12+ab}=\\
\frac{144(ab+ac+bc)+12(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+12abc}{1728abc+144(a^2b^2+b^2c^2+a^2c^2)+12abc(a^2+b^2+c^2)+(abc)^2}\\
Fazendo:\\
X_1=ab+ac+bc\\
X_2 = a^2b+ab^2+a^2c+ac^2+b^2c+bc^2\\
X_3 = abc\\
X_4=a^2+b^2+c^2\\
X_5=a^2b^2+b^2c^2+a^2c^2\\
X_2=a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=
a^2(b+c)+b^2(a+c)+c^2(a+b)=\\a^2(12−a)+b^2(12−b)+c^2(12−c)=
12(a^2+b^2+c^2)−(a^3+b^3+c^3)=12\cdot X_4−1440(I)\\
144=12^2=(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)=
X_4+2_X1⟹\boxed{\mathsf{X_4=144−2X_1}}(II)\\
123−1440=(a+b+c)^3−(a^3+b^3+c^3)=
3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc=3X_2+6X_3⟹
96=X_2+2X_3⟹X_2=96−2X_3(III)\\
X_1^2=(ab+ac+bc)^2=(a^2b^2+a^2c^2+b^2c^2)+2abc(a+b+c)=
X_5+24X_3⟹X_5=X_1^2−24X_3(IV)\\
I~e~II: X_2=12X_4-1440=12(144-2X_1)-1440\rightarrow \boxed{\mathsf{X_2=288-24X_1}}(IV)\\
III~e~IV:96−2X_3=X_2=288−24X_1⟹24X_1=192+2X_3 \rightarrow \boxed{\mathsf{X_3=12X_1−96}}(V)\\
IV~e~V: X_5=X_1^2−24X_3=X1^2−24(12X_1−96)\rightarrow \boxed{\mathsf{X_5=X_1^2−288X_1+2304}}\\
Substituindo X_1, X_2, X_3, X_4, X_5: \frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12+ab}=\frac{144X_1+12X_2+12X_3}{1728X_3+144X_5+12X_3X_4+X_3^2}\\
=\frac{144X_1+12(288-24X_1)+12(12X_1-96)}{1728(12X_1-96)+144(X_1^2-288X_1+2304)+12(12X_1-96)(144-2X_1)+(12X_1-96)^2}=\\
\frac{2304}{9216}=\boxed{\boxed{\mathsf{\frac{1}{4}}}}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{\frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12+ab}=\\
\frac{144(ab+ac+bc)+12(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+12abc}{1728abc+144(a^2b^2+b^2c^2+a^2c^2)+12abc(a^2+b^2+c^2)+(abc)^2}\\
Fazendo:\\
X_1=ab+ac+bc\\
X_2 = a^2b+ab^2+a^2c+ac^2+b^2c+bc^2\\
X_3 = abc\\
X_4=a^2+b^2+c^2\\
X_5=a^2b^2+b^2c^2+a^2c^2\\
X_2=a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=
a^2(b+c)+b^2(a+c)+c^2(a+b)=\\a^2(12−a)+b^2(12−b)+c^2(12−c)=
12(a^2+b^2+c^2)−(a^3+b^3+c^3)=12\cdot X_4−1440(I)\\
144=12^2=(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)=
X_4+2_X1⟹\boxed{\mathsf{X_4=144−2X_1}}(II)\\
123−1440=(a+b+c)^3−(a^3+b^3+c^3)=
3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc=3X_2+6X_3⟹
96=X_2+2X_3⟹X_2=96−2X_3(III)\\
X_1^2=(ab+ac+bc)^2=(a^2b^2+a^2c^2+b^2c^2)+2abc(a+b+c)=
X_5+24X_3⟹X_5=X_1^2−24X_3(IV)\\
I~e~II: X_2=12X_4-1440=12(144-2X_1)-1440\rightarrow \boxed{\mathsf{X_2=288-24X_1}}(IV)\\
III~e~IV:96−2X_3=X_2=288−24X_1⟹24X_1=192+2X_3 \rightarrow \boxed{\mathsf{X_3=12X_1−96}}(V)\\
IV~e~V: X_5=X_1^2−24X_3=X1^2−24(12X_1−96)\rightarrow \boxed{\mathsf{X_5=X_1^2−288X_1+2304}}\\
Substituindo X_1, X_2, X_3, X_4, X_5: \frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12+ab}=\frac{144X_1+12X_2+12X_3}{1728X_3+144X_5+12X_3X_4+X_3^2}\\
=\frac{144X_1+12(288-24X_1)+12(12X_1-96)}{1728(12X_1-96)+144(X_1^2-288X_1+2304)+12(12X_1-96)(144-2X_1)+(12X_1-96)^2}=\\
\frac{2304}{9216}=\boxed{\boxed{\mathsf{\frac{1}{4}}}}}[/tex3]

[petras]

Outra solução do colega Ragu:

[tex3]\mathsf{ a+b+c=12(E01)\\
a^2+b^2+c^2=(a+b+c)^2−2(ab+bc+ca)=144−2(ab+bc+ca) (E02)\\
a^3+b^3+c^3=1440\\
⟹1440=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)+3abc=12(144−3(ab+bc+ca))+3abc\\

Dividindo~ por~ 3:\\

⟹480=576−12(ab+bc+ca)+abc⟹12(ab+bc+ca)−abc=96(E03)\\
Substituindo ~12~ por~(a+b+c)~ em~ \frac{1}{12a+bc}:\\
\frac{1}{12a+bc}=\frac{1}{(a+b+c)a+bc}=\frac{1}{a^2+ab+ac+bc}=\frac{1}{(a+c)(a+b)}\\
Do~ mesmo~ modo, \frac{1}{12b+ac}=\frac{1}{(b+c)(a+b)}\\
\frac{1}{12c+ab}=\frac{1}{(b+c)(c+a)}\\
\frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12c+ab}=\frac{1}{(b+c)(a+b)+1(b+c)(a+b)}+\frac{1}{(b+c)(c+a)}\\
=\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}=\frac{24}{(a+b)(b+c)(c+a)}(E04)\\
Expandindo~ o~ denominador~ de~ (E04):\\
(a+b)(b+c)(c+a)=(ab+ac+b^2+bc)(c+a)=abc+a^2b+ac^2+a^2c+b^2c+b^2c+bc^2+abc \\
=2abc+ab(a+b)+bc(b+c)+ca(c+a)=2abc+ab(12−c)+bc(12−a)+ca(12−b)\\
=2abc+12(ab+bc+ca)−3abc=\overbrace{12ab+12bc+12ca−abc} (de(E03)) = 96(E05)\\

Substitutindo~ (E05)~em~ (E04):
\frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12c+ab}=\frac{24}{96}=\boxed{\boxed{\mathsf{\frac{1}{4}}}}} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{ a+b+c=12(E01)\\
a^2+b^2+c^2=(a+b+c)^2−2(ab+bc+ca)=144−2(ab+bc+ca) (E02)\\
a^3+b^3+c^3=1440\\
⟹1440=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)+3abc=12(144−3(ab+bc+ca))+3abc\\

Dividindo~ por~ 3:\\

⟹480=576−12(ab+bc+ca)+abc⟹12(ab+bc+ca)−abc=96(E03)\\
Substituindo ~12~ por~(a+b+c)~ em~ \frac{1}{12a+bc}:\\
\frac{1}{12a+bc}=\frac{1}{(a+b+c)a+bc}=\frac{1}{a^2+ab+ac+bc}=\frac{1}{(a+c)(a+b)}\\
Do~ mesmo~ modo, \frac{1}{12b+ac}=\frac{1}{(b+c)(a+b)}\\
\frac{1}{12c+ab}=\frac{1}{(b+c)(c+a)}\\
\frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12c+ab}=\frac{1}{(b+c)(a+b)+1(b+c)(a+b)}+\frac{1}{(b+c)(c+a)}\\
=\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}=\frac{24}{(a+b)(b+c)(c+a)}(E04)\\
Expandindo~ o~ denominador~ de~ (E04):\\
(a+b)(b+c)(c+a)=(ab+ac+b^2+bc)(c+a)=abc+a^2b+ac^2+a^2c+b^2c+b^2c+bc^2+abc \\
=2abc+ab(a+b)+bc(b+c)+ca(c+a)=2abc+ab(12−c)+bc(12−a)+ca(12−b)\\
=2abc+12(ab+bc+ca)−3abc=\overbrace{12ab+12bc+12ca−abc} (de(E03)) = 96(E05)\\

Substitutindo~ (E05)~em~ (E04):
\frac{1}{12a+bc}+\frac{1}{12b+ac}+\frac{1}{12c+ab}=\frac{24}{96}=\boxed{\boxed{\mathsf{\frac{1}{4}}}}} [/tex3]