Questão 1)
A racionalização consiste em tranformar a fração com denominador irracional em racional através da multiplicação por um radical
Letra A)
[tex]\begin{array}{l}\sf \dfrac{10}{\sqrt{3}}=\\\\\sf =\dfrac{10}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}\\\\\!\boxed{\sf =\dfrac{10\sqrt{3}}{3}} \\ \\ \end{array}[/tex]
Letra B)
[tex]\begin{array}{l}\sf \dfrac{1}{\sqrt{23}}=\\\\\sf =\dfrac{1}{\sqrt{23}}\cdot\dfrac{\sqrt{23}}{\sqrt{23}}\\\\\boxed{\sf =\dfrac{\sqrt{23}}{23}}\\\\\end{array}[/tex]
Letra C)
[tex]\begin{array}{l}\sf \dfrac{8}{\sqrt{8}}=\\\\\sf =\dfrac{8}{2\sqrt{2}}\\\\\sf =\dfrac{4}{\sqrt{2}}\\\\\sf =\dfrac{4}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf =\dfrac{4\sqrt{2}}{2}\\\\\!\boxed{\sf =2\sqrt{2}}\\\\\end{array}[/tex]
Letra D)
[tex]\begin{array}{l}\sf \dfrac{5}{\sqrt[\sf3]{\sf7^2}}=\\\\\sf =\dfrac{5}{\sqrt[\sf3]{\sf7^2}}\cdot\dfrac{\sqrt[\sf3]{\sf7}}{\sqrt[\sf3]{\sf7}}\\\\\sf =\dfrac{5\sqrt[\sf3]{\sf7}}{\sqrt[\sf3]{\sf7^3}}\\\\\!\boxed{\sf =\dfrac{5\sqrt[\sf3]{\sf7}}{7}}\\\\\end{array}[/tex]
Letra E)
[tex]\begin{array}{l}\sf \dfrac{3}{\sqrt[\sf5]{2^2}}=\\\\\sf =\dfrac{3}{\sqrt[\sf5]{2^2}}\cdot\dfrac{\sqrt[\sf5]{2^3}}{\sqrt[\sf5]{2^3}}\\\\\sf = \dfrac{3\sqrt[\sf5]{2^3}}{\sf\sqrt[\sf5]{2^5}}\\\\\!\boxed{\sf = \dfrac{3\sqrt[\sf5]{8}}{2}}\\\\\end{array}[/tex]
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Questão 2)
Para determinar se a equação terá raízes reais a partir do valor do delta, veja as regras a seguir
[tex] \small\begin{array}{l}~~~\bullet~\sf Se~\:\Delta > 0~\to~x'~e~\:x''\!\in\mathbb{R},~\:com\:~x'\,\neq\, x''\\\\~~~\bullet~\sf Se~\:\Delta = 0~\to~x'~e~\:x''\!\in\mathbb{R},~\:com\:~x'= x''\\\\~~~\bullet~\sf Se~\:\Delta < 0~\to~x'~e~\:x''\!\notin\mathbb{R}\end{array}[/tex]
Assim:
Letra A)
[tex]\begin{array}{l}\sf 2x^2-3x+1=0\\\\\sf \Delta=b^2-4ac\\\\\sf \Delta=(-3)^2-4(2)(1)\\\\\sf \Delta=9-8\\\\\!\boxed{\sf \Delta=1}\sf~\to~\Delta > 0\end{array}[/tex]
Assim a equação terá raízes reais
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Letra B)
[tex]\begin{array}{l}\sf -3x^2+10x-3=0\\\\\sf \Delta=b^2-4ac\\\\\sf \Delta=(10)^2-4(-3)(-3)\\\\\sf \Delta=100-36\\\\\!\boxed{\sf \Delta=64}\sf~\to~\Delta > 0\end{array}[/tex]
Assim a equação terá raízes reais
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Letra C)
[tex]\begin{array}{l}\sf 2x^2+3x+25=0\\\\\sf \Delta=b^2-4ac\\\\\sf \Delta=(3)^2-4(2)(25)\\\\\sf \Delta=9-200\\\\\!\boxed{\sf \Delta=-191}\sf~\to~\Delta < 0~\therefore~x\notin\mathbb{R}\end{array}[/tex]
Assim a equação não terá raízes reais
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Questão 3)
Usando as fórmulas de soma e produto:
Letra A)
[tex]\begin{array}{l}\sf 4x^2+x-14=0\\\\\sf S=-\dfrac{b}{a}\quad e\quad P=\dfrac{c}{a}\\\\\sf S=-\dfrac{1}{4}\quad e\quad P=\dfrac{-14~}{4}\\\\\!\boxed{\sf S=-\dfrac{1}{4}\quad e\quad P=-\dfrac{7}{2}} \\ \\ \end{array}[/tex]
Letra B)
[tex]\begin{array}{l}\sf -9x^2+x-8=0\\\\\sf S=-\dfrac{b}{a}\quad e\quad P=\dfrac{c}{a}\\\\\sf S=-\dfrac{1}{-9~}\quad e\quad P=\dfrac{-8~}{-9~}\\\\\!\boxed{\sf S=\dfrac{1}{9}\quad e\quad P=\dfrac{8}{9}}\\\\\end{array}[/tex]
Letra C)
[tex]\begin{array}{l}\sf x^2+4x-5=0\\\\\sf S=-\dfrac{b}{a}\quad e\quad P=\dfrac{c}{a}\\\\\sf S=-\dfrac{4}{1}\quad e\quad P=\dfrac{-5~}{1}\\\\\!\boxed{\sf S=-4\quad e\quad P=-5}\end{array}[/tex]
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Questão 4)
Para identificar os coeficientes, lembre-se da forma geral:
ax² + bx + c = 0
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Letra A)
x² – 7x + 10 = 0
a = 1, b = – 7, c = 10
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Letra B)
x²– x – 6 = 0
a = 1, b = – 1, c = – 6
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Letra C)
4x² – 4x + 1 = 0
a = 4, b = – 4, c = 1
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Letra D)
– x² – 5x – 1 = 0
a = – 1, b = – 5, c = – 1
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Letra E)
4x² = 0
a = 4, b = 0, c = 0
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Letra F)
3x² – 7 = 0
a = 3, b = 0, c = – 7
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Letra G)
x²– 2x = 0
a = 1, b = – 2, c = 0
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Letra H)
12x² + 4 – 5x = 0
a = 12, b = – 5, c = 4
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Questão 5)
Resolvendo as equações incompletas:
Letra A)
[tex]\begin{array}{l}\sf x^2-8x=0\\\\\sf x(x-8)=0\\\\ \begin{cases}\sf x=0\\\\\sf x-8=0\end{cases}\\\\\begin{cases}\sf x'=0\\\\\sf x''=8\end{cases}\\\\\sf S=\Big\{0~~;~~8\Big\} \\ \\ \end{array}[/tex]
Letra B)
[tex]\begin{array}{l}\sf x^2+3x=0\\\\\sf x(x+3)=0\\\\ \begin{cases}\sf x=0\\\\\sf x+3=0\end{cases}\\\\\begin{cases}\sf x'=0\\\\\sf x''=-3\end{cases}\\\\\sf S=\Big\{-3~~;~~0\Big\}\\\\\end{array}[/tex]
Letra C)
[tex]\begin{array}{l}\sf 9x^2-72=0\\\\\sf 72+9x^2-72=0+72\\\\\sf 9x^2=72\\\\\sf \dfrac{9x^2}{9}=\dfrac{72}{9}\\\\\sf x^2=8\\\\\sf \sqrt{x^2}=\sqrt{8}\\\\\sf |x|=\sqrt{8}\\\\\sf x=\pm~\sqrt{8}\\\\\sf x=\pm~2\sqrt{2}\\\\\sf S=\Big\{-2\sqrt{2}~~;~~2\sqrt{2}\Big\}\\\\\end{array}[/tex]
Letra D)
[tex]\begin{array}{l}\sf 5x^2-20=0\\\\\sf 20+5x^2-20=0+20\\\\\sf 5x^2=20\\\\\sf \dfrac{5x^2}{5}=\dfrac{20}{5}\\\\\sf x^2=4\\\\\sf \sqrt{x^2}=\sqrt{4}\\\\\sf |x|=2\\\\\sf x=\pm~2\\\\\sf S=\Big\{-2~~;~~2\Big\}\\\\\end{array}[/tex]
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Questão 6)
Se imaginarmos, vemos um triângulo retângulo, onde temos que os catetos que são 15 m e 8 m, e a hipotenusa que é a altura da escada ( chamaremos de " h " )
Dessa forma aplicando o Teorema de Pitágoras:
[tex]\begin{array}{l}\sf a^2=b^2+c^2\\\\\sf (h)^2=(15)^2+(8)^2\\\\\sf h^2=225+64\\\\\sf h^2=289\\\\\sf \sqrt{h^2}=\sqrt{289}\\\\\!\boxed{\sf h=17} \\ \\ \end{array}[/tex]
Assim, resposta é LETRA D) 17 m
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Att. Nasgovaskov
