Question

Resolva as equações abaixo

d) (x – ½)² = 1/16

e) (x + √2)² = 8

f) (x – 3/2)² = ¼

me ajudem pra agora Pfv ​

Answer 1

Explicação passo a passo:

d)

(x – ½)² = 1/16

quadrado da diferença conforme regra

[( x )² - 2 * x * 1/2 + ( 1/2)² ]

x² - 2x/2 + 1/4 =

2x/2 = corta 2 = x

x² - x + 1/4 = 1/16

x²/1 - 1x/1 + 1/4 = 1/16

mmc = 16

16 : 1 = 16 * x² = 16x² >>>

16 ; 1 = 16x * 1x = 16x >>>>>

16 : 4 = 4 * 1 = 4 >>>>

16 : 16 = 1 * 1 = 1 >>>>

16x² - 16x + 4 = 1

16x² - 16x + 4 - 1 = 0

16x² - 16x +3 = 0 >>>>>>>>>>>resposta

e)

(x + √2)² = 8

quadrado da soma conforme regra

[ ( x)² +2 * x * V2 + ( V2)² ] =8

x² + 2x.V2 + 2 = 8

x² + 2x.V2 + 2 - 8 =0

X² + 2X.V2 - 6 = 0 RESPOSTA

f)

(x – 3/2)² = ¼

3/2 = 3 ; 2 = 1,5 >>>>

1/4 = 1 : 4 =0,25

( X - 1,5)² = 0,25

[ (X)² - 2 *X * 1,5 + ( 1,5)² ] = 0,25

X² - 3X + 2,25 = 0,25

X² - 3X + 2,25 - 0,25 = 0

X² - 3X +2 >>>>>RESPOSTA

Answer 2

Resposta:

$d)\\ \\ (x-\dfrac{1}{2})^2=\dfrac{1}{16}\\ \\ \\ (x-\dfrac{1}{2})=\pm\sqrt{\dfrac{1}{16}} \\ \\ \\ x-\dfrac{1}{2}=\pm\dfrac{1}{4}\\ \\ \\ x-\dfrac{1}{2}=\dfrac{1}{4}\\ \\\\ x'=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1+2}{4}=\boxed{\dfrac{3}{4}}\\ \\ \\ x"-\dfrac{1}{2}=-\dfrac{1}{4}\\ \\ \\ x"=-\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{-1+2}{4}=\boxed{\dfrac{1}{4}}$

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e)

$(x+\sqrt{2} )^2=8\\ \\ (x+\sqrt{2} )=\pm\sqrt{8} \\ \\ (x+\sqrt{2} )=\pm2\sqrt{2} \\ \\ \\ x+\sqrt{2} =2\sqrt{2} \\ \\ x'=2\sqrt{2} -\sqrt{2} \\ \\ \boxed{x'=\sqrt{2} }\\ \\ \\ x+\sqrt{2} =-2\sqrt{2} \\ \\ x"=-2\sqrt{2} -\sqrt{2} \\ \\\boxed{ x"=-3\sqrt{2} }$

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f)

$(x-\dfrac{3}{2})^2=\dfrac{1}{4}\\ \\ \\ (x-\dfrac{3}{2})=\pm\sqrt{\dfrac{1}{4}} \\ \\ \\ x-\dfrac{3}{2}=\pm\dfrac{1}{2}\\ \\ \\ x'-\dfrac{3}{2}=\dfrac{1}{2}\\ \\ \\ x'=\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{1+3}{2}=\dfrac{4}{2}=\boxed{2}\\ \\ \\ x"=-\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{2}{2}=\boxed{1}$