Question

URGENTEEEEEE !

Nas expressões, introduza o fator externo no radicando:

a) 2√2

b) 5√3

c) 2∛10

d) 3 vezes raiz quarta de 2

e) 2∛7

f) 6√2

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Resolva os sistemas

a) {x²+y²=13
    {x-y=1

b) {x/y=1/2
    {x²+y=35

c) {x-2y=-2
    {x²+xy=8

Answer 1

a)
$2 \sqrt{2} = \sqrt{2.2^2} = \sqrt{2.4} = \sqrt{8}$

b)
$5 \sqrt{3} = \sqrt{3.5^2} = \sqrt{3.25} = \sqrt{75}$

c)
$2 \sqrt[3]{10} = \sqrt[3]{10.2^3} = \sqrt[3]{10.8} = \sqrt[3]{80}$

d)
$3 \sqrt[4]{2} = \sqrt[4]{2.3^4} = \sqrt[4]{2.81} = \sqrt[4]{162}$

e)
$2 \sqrt[3]{7} = \sqrt[3]{7.2^3} = \sqrt[3]{7.8} = \sqrt[3]{56}$

f)
$6 \sqrt{2} = \sqrt{2.6^2} = \sqrt{2.36} = \sqrt{72}$

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Sistemas

a)
$\left \{ {{x^2+y^2=13} \atop {x-y=1}} \right.$

isolar  (substituição)
x=1+y

subst/ em
x²+y²=13

(1+y)²+y²=13
1+2y+y²+y²=13
2y²+2y+1-13=0
2y²+2y-12=0  (÷2)
y²+y-6=0

Δ=b²-4ac
Δ=1+24
Δ=25
√Δ=√25=± 5

$y= \frac{-1\pm5}{2}$

$y'= \frac{-1-5}{2} =- \frac{6}{2} =-3$

$y"= \frac{-1+5}{2} = \frac{4}{2} =2$

como
x=1+y                        x=1+y            
p/ y'=-3                        p/ y"=2
x'=1-3                          x"=1+2
x'=-2                            x"=3

S={(-2,-3);(3,2)}

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b)
$\left \{ {{ \frac{x}{y} = \frac{1}{2} } \atop {x^2+y=35}} \right.$

y=2x

subst/ em

x²+y=35
x²+2x=35
x²+2x-35=0

Δ=b²-4ac
Δ=4+140
Δ=144
√Δ=√144=± 12

$x= \frac{-2\pm12}{2}$

$x'= \frac{-2+12}{2} = \frac{10}{2} =5$

$x"= \frac{-2-12}{2} =- \frac{14}{2} =-7$

como
y=2x                                  y=2x
p/x'=5                                 p/ x"=-7
y'=2(5)                                  y"=2(-7)
y'=10                                    y"=-14

S={(5,10);(-7,-14)}

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c)
$\left \{ {{x-2y=-2} \atop {x^2+xy=8}} \right.$

isola
x=2y-2

subst/  em
x²+xy=8
(2y-2)²+y(2y-2)=8
4y²-8y+4+2y²-2y-8=0
6y²-10y-4=0    (÷2)
3y²-5y-2=0

Δ=b²-4ac
Δ=25+24
Δ=49
√Δ=√49=± 7

$y= \frac{5\pm7}{6}$

$y'= \frac{5+7}{6} = \frac{12}{6} =2$

$y"= \frac{5-7}{6} =- \frac{2}{6} =- \frac{1}{3}$

como
x=2y-2
p/y'=$- \frac{1}{3}$

$x'=2( \frac{-1}{3} )-2$

$x'=- \frac{2}{3} -2= \frac{-2-6}{3} =- \frac{8}{3}$

p/y"=2
x"=2(2)-2
x"=4-2
x"=2

S={($- \frac{8}{3} ,- \frac{1}{3}$);(2,2)}