Question

URGEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEENTE !

Resolva as expressões algébricas apresentadas a seguir:
A) 4 -(x-5)²= 2(x-3)(x+3)

B) 3(x-1) - x = x (x-2)
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4 2

C) 2 ³RAIZ DE 3X-2=5


D)X+3RAIZ DE X+4 = 2 X +4

Answer 1

a) $4 - (x-5)^2 = 2(x-3)(x+3) \\ 4-(x^2-10x+25)=2x^2-18 \\ 3x^2-10x+3=0 \\ \\ x'= \frac{10+ \sqrt{100-4*3*3} }{12} = \frac{10+ \sqrt{64} }{12} = \frac{18}{12} = \frac{3}{4} \\ x''= \frac{10- \sqrt{100-4*3*3} }{12} = \frac{2}{12} = \frac{1}{6}$
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b) $\frac{3(x-1)}{4}-x= \frac{x(x-2)}{2} \\ \frac{3(x-1)-4x=2x(x-2)}{4} \\ 3x-3-4x=2x^2-4x \\ 2x^2-3x+3=0 \\ \\ x'= \frac{3+ \sqrt{9-4*2*3} }{4}= \frac{3+ \sqrt{-15} }{12} = \frac{3+i \sqrt{15} }{12} \\ x''= \frac{3- \sqrt{9-4*2*3} }{4}= \frac{3- \sqrt{-15} }{12} = \frac{3-i \sqrt{15} }{12}$

Com: $i = \sqrt{-1}$
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c) $2 \sqrt[3]{3x-2} = 5 \\ (2 \sqrt[3]{3x-2})^3 = 5^3 \\ 8(3x-2)=125 \\ 24x=141 \\ x= \frac{141}{24} = \frac{47}{8}$
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d) $x+3 \sqrt{x+4}=2x+4 \\ 3 \sqrt{x+4}=x+4 \\ (3 \sqrt{x+4})^2=(x+4)^2 \\ 3(x+4)=(x+4)(x+4) \\ 3=x+4 \\ x=-1$