Question

resolva as equações.

a ) √x + 2 = 7
b ) √3x + 1 + 6 = 2
c ) √5x - 10 = √3x + 2
d ) 3√3x + 1 = √18
e ) √3 + 6 - 2 = x

Answer 1

Olá!

a)

$\sqrt{x + 2} = 7 \\ ( \sqrt{x + 2 } )^{2} = {7}^{2} \\ x + 2 = 49 \\ x = 49 - 2 \\ x = 47$

b)

$\sqrt{3x + 1 + 6} = 2 \\ ( \sqrt{3x + 1 +6 } ) ^{2} = {2}^{2} \\ 3x + 1 + 6 = 4 \\ 3x = 4 - 1 - 6 \\ 3x = - 3 \\ x = - 3 \div 3 \\ x = - 1$

c)

$\sqrt{5x - 10} = \sqrt{3x + 2} \\ (\sqrt{5x - 10} ) ^{2} = (\sqrt{3x + 2}) ^{2} \\ 5x - 10 = 3x + 2 \\ 5x - 3x = 2 + 10 \\ 2x = 12 \\ x = 12 \div 2 \\ x = 6$

d)

$3 \sqrt{3x + 1} = \sqrt{18}$

√18 = 3√2

$3 \sqrt{3x + 1} = 3 \sqrt{2}$

Divide ambos por 3:


$\sqrt{3x + 1} = \sqrt{2} \\ ( \sqrt{3x + 1} ) ^{2} = ({ \sqrt{2} }) ^{2} \\ 3x + 1 = 2 \\ 3x = 2 - 1 \\ 3x = 1 \\ x = \frac{1}{3}$

e)


$\sqrt{3 + 6 - 2} = x \\ ({ \sqrt{3 + 6 - 2} })^{2} = {x}^{2} \\ 3 + 6 - 2 = {x}^{2} \\ 7 = {x}^{2} \\ x = \sqrt{7}$