Question

Usando o Teorema de Tales, resolva os exercícios abaixo.​

Answer 1

Explicação passo-a-passo:

a)

Pelo Teorema de Tales:

$\sf \dfrac{2}{4}=\dfrac{3}{x}$

$\sf 2x=4\cdot3$

$\sf 2x=12$

$\sf x=\dfrac{12}{2}$

$\sf \red{x=6}$

b)

Pelo Teorema de Tales:

$\sf \dfrac{4}{12}=\dfrac{x}{21}$

$\sf 12x=4\cdot21$

$\sf 12x=84$

$\sf x=\dfrac{84}{12}$

$\sf \red{x=7}$

c)

Pelo Teorema de Tales:

$\sf \dfrac{x}{14}=\dfrac{9}{12}$

$\sf 12x=14\cdot9$

$\sf 12x=126$

$\sf x=\dfrac{126}{12}$

$\sf \red{x=10,5}$

d)

Pelo Teorema de Tales:

$\sf \dfrac{x}{9}=\dfrac{8}{12}$

$\sf 12x=9\cdot8$

$\sf 12x=72$

$\sf x=\dfrac{72}{12}$

$\sf \red{x=6}$

e)

Pelo Teorema de Tales:

$\sf \dfrac{9}{x}=\dfrac{18}{4}$

$\sf 18x=9\cdot4$

$\sf 18x=36$

$\sf x=\dfrac{36}{18}$

$\sf \red{x=2}$

f)

Pelo Teorema de Tales:

$\sf \dfrac{10}{15}=\dfrac{3x+1}{5x-2}$

$\sf 10\cdot(5x-2)=15\cdot(3x+1)$

$\sf 50x-20=45x+15$

$\sf 50x-45x=15+20$

$\sf 5x=35$

$\sf x=\dfrac{35}{5}$

$\sf \red{x=7}$

g)

Pelo Teorema de Tales:

$\sf \dfrac{x-5}{x+1}=\dfrac{6}{24}$

$\sf 24\cdot(x-5)=6\cdot(x+1)$

$\sf 24x-120=6x+6$

$\sf 24x-6x=6+120$

$\sf 18x=126$

$\sf x=\dfrac{126}{18}$

$\sf \red{x=7}$

h)

Pelo Teorema de Tales:

$\sf \dfrac{8}{x+1}=\dfrac{12}{2x-6}$

$\sf 8\cdot(2x-6)=12\cdot(x+1)$

$\sf 16x-48=12x+12$

$\sf 16x-12x=12+48$

$\sf 4x=60$

$\sf x=\dfrac{60}{4}$

$\sf \red{x=15}$

Answer 2

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$\tt a)~\sf \dfrac{2}{4}=\dfrac{3}{x}\\\sf 2x=12\\\sf x=\dfrac{12}{2}\\\sf x=6\bf\checkmark$

$\tt b)~\sf\dfrac{4}{\diagup\!\!\!\!12_4}=\dfrac{x}{\diagup\!\!\!\!\!21_7}\\\sf\dfrac{4}{4}=\dfrac{x}{7}\\\sf 1=\dfrac{x}{7}\\\sf x=7\checkmark$

$\tt c)~\sf\dfrac{x}{14}=\dfrac{\diagup\!\!\!9^3}{\diagup\!\!\!\!\!\!12_4}\\\sf \dfrac{x}{14}=\dfrac{3}{4}\\\sf x=\dfrac{\diagup\!\!\!\!\!14^7\cdot3}{\diagup\!\!\!\!4_2}\\\sf x=\dfrac{21}{2}\checkmark$

$\tt d)~\sf\dfrac{x}{12}=\dfrac{8}{9}\\\sf x=\dfrac{\diagup\!\!\!\!\!12^4\cdot8}{\diagup\!\!\!9_3}\\\sf x=\dfrac{32}{3}\checkmark$

$\tt e)~\sf\dfrac{x}{4}=\dfrac{\diagup\!\!\!9^1}{\diagup\!\!\!\!\!18_2}\\\sf\dfrac{x}{4}=\dfrac{1}{2}\\\sf x=\dfrac{4\cdot1}{2}=\dfrac{4}{2}=2\checkmark$

$\tt f)~\sf\dfrac{5x-2}{\diagup\!\!\!\!\!10_2}=\dfrac{3x+1}{\diagup\!\!\!\!\!15_3}\\\sf\dfrac{5x-2}{2}=\dfrac{3x+1}{3}\\\sf15x-6=6x+2\\\sf15x-6x=6+2\\\sf9x=8\\\sf x=\dfrac{8}{9}\checkmark$

$\tt g)~\sf\dfrac{x-5}{x+1}=\dfrac{\diagup\!\!\!\!6^1}{\diagup\!\!\!\!\!24_4}\\\sf 4\cdot(x-5)=1\cdot(x+1)\\\sf 4x-20=x+1\\\sf 4x-x=20+1\\\sf 3x=21\\\sf x=\dfrac{21}{3}\\\sf x=7\checkmark$

$\tt h)~\sf\dfrac{2x-6}{\diagup\!\!\!\!\!12_3}=\dfrac{x+1}{\diagup\!\!\!\!8_2}\\\sf 2\cdot(2x-6)=3\cdot(x+1)\\\sf 4x-12=3x+3\\\sf 4x-3x=12+3\\\sf x=15\checkmark$