Matemática, 08.03.2020 04:50 superlucas007

3x=2(x-1)+3
Me ajudem por favor

Respostas: 3

Mostrar respostas

Respostas

Resposta de: alvesdasilvamariaedu

resposta:

3x=2x-2+3

3x=2x+1

3x-2x=1

x=1

Resposta de: julliagatinhappan90k

3x = 2(x-1) + 3

3x = 2x - 2 + 3

3x -2x = -2 + 3

x = 1

primeiro voce faz a distributiva 2(x-1), multiplicando o x e o -1 por 2.

depois resolve o resto

Resposta de: alves1250

3x=2(x-1)² +3
3x=2(x²-2x+1)+3
3x=2x² -4x + 2 +3
-2x²+4x+3x -5 =0
-2x² +7x-5=0 (-1)
2x² -7x +5 =0
Báscara:
x= (-(-7) +/- √(-7)² - 4.2.5) / 2.2

x= (7 +/- √49 -40 ) / 4

x= (7 +/- √9) / 4
x= (7 +/- 3) / 4
x1= 10/4 = 5/2
x2=4/4 = 1

Solução S = {5/2 , 1}

Resposta de: yarawaneska49

3x² = 2 ( x -1)² + 3
3x² = 2.(x - 1).(x - 1) + 3
3x² = 2.(x² - x - x + 1) + 3
3x² = 2x² - 2x - 2x + 2 + 3
3x² - 2x² + 2x + 2x - 2 - 3 = 0
x² + 4x - 5 = 0

Resolve-se por Bháskara:

∆ = b² - 4.a.c
∆ = 4² - 4.1.(-5)
∆ = 16 + 20
∆ = 36

x = [-b ± √∆] / (2.a)
x = [-(4) ±√36] / (2.1)
x = [-4 ± 6] / 2
x1 = [-4 + 6] / 2 → x1 = 2 / 2 → x1 = 1
x2 = [-4 - 6] / 2 → x2 = (-10) / 2 → x2 = -5

S = {-5; 1}

Resposta de: carlosmucuri20pbd05m

3x² = 2[(x -1) , (x -1) ]+ 3
3x² -3 = 2 . (x² -2x +1)
3x² -3 -2x² +4x -2 = 0
x² +4x -5 = 0

Δ = 16 -4 . -5
Δ = 36

x = (-4 ± 6)/2

x' = 1
x'' = -5

S = {1 , -5}

Resposta de: jvoliveiraneonp55fex

$3x^2=2(x-1)^2+3 \\
3x^2=2(x^2-2x+1)+3 \\
3x^2=2x^2-4x+2+3 \\
x^2+4x-5=0 \\
\Delta=4^4-4.1.(-5)=16+20=36 \\
\\
x=\frac{-4+-\sqrt{36}}{2}=\frac{-4+-6}{2} \\
\\
x_1=-5 \\
\\
x_2=1$

$x(x+11)+2(x+21)=0 \\
x^2+11x+2x+42=0 \\
x^2+13x+42=0 \\
\Delta=13^2-4.1.42=169-168=1 \\
\\
x=\frac{-13+-1}{2} \\
\\
x_1=-7 \\
\\
x_2=-6$

$6(x^2-1)-14=5x^2+x \\
6x^2-6-14-5x^2-x=0 \\
x^2-x-20=0 \\
\\
\Delta=(-1)^2-4.1.(-20)=1+80=81 \\
\\
x=\frac{1+-\sqrt{81}}{2}=\frac{1+-9}{2} \\
\\
x_1=-4 \\
\\
x_2=5$

Resposta de: pattiniantonia92

$a) \\ (x+2)^2+x=0 \\ \\ (x + 2)(x+ 2) + x \\ \\ x^2 + 2x + 2x + 4 + x = 0 \\ \\ x^2 + 5x + 4 = 0$

Resolvendo por Bháskara:

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=5,c=4
Δ= b^2 - 4ac
Δ= 5^2 - 4*1*4
Δ= 25 - 16
Δ= 9

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-5 \pm \sqrt{9}}{2*1} \\ \\ \\ x = \dfrac{-5 \pm 3}{2} \\ \\ \\ x' = \dfrac{-5 + 3}{2} \\ \\ \\ x' = \dfrac{-2}{2} \\ \\ \\ x' = -1 \\ \\ \\ x'' = \dfrac{-5 - 3}{2} \\ \\ \\ x'' = \dfrac{-8}{2} \\ \\ \\ x'' = -4$

S = {-1, -4}

==============================================

$b) \\ 3x=2(x-1)+3 \\ \\ 3x = 2x - 2 + 3 \\ \\ 3x - 2x - 2 + 3 = 0$

==============================================
$c) \\ x(x+11)+2(x+21)=0 \\ \\ x^2 + 11x + 2x + 42 \\ \\ x^2 + 13x + 42 = 0$

a=1, b=13,c=42
Δ= b^2 - 4ac
Δ= 13^2 - 4*1*42
Δ= 169 - 168
Δ= 1

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-13 \pm \sqrt{1}}{2*1} \\ \\ \\ x = \dfrac{-13 \pm 1}{2} \\ \\ \\ x' = \dfrac{-13 + 1}{2} \\ \\ \\ x' = \dfrac{-12}{2} \\ \\ \\ \\ x' = -6 \\ \\ \\ x'' = \dfrac{-13 - 1}{2} \\ \\ \\ x'' = \dfrac{-14}{2} \\ \\ \\ x'' = -7
$

S = {-6, -7}
==============================================
$d) \\ \\ 6(x^2-1)-14=5x^2+x \\ \\ 6x^2 - 6 - 14 = 5x^2 + x \\ \\ 6x^2 -20 = 5x^2 + x \\ \\ x^2 -x - 20 = 0$
$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

a=1, b=-1,c=-20
Δ= b^2 - 4ac
Δ= (-1)^2 - 4*1*-20
Δ= 1 + 80
Δ= 81

$x = \dfrac{-(-1) \pm \sqrt{81}}{2*1} \\ \\ \\ x = \dfrac{1 \pm 9}{2} \\ \\ \\ x' = \dfrac{1 + 9}{2} \\ \\ \\ x' = \dfrac{10}{2} \\ \\ \\ x' = 5 \\ \\ \\ x'' = \dfrac{1 - 9}{2} \\ \\ \\ x'' = \dfrac{-8}{2} \\ \\ \\ x'' = -4$

S = {5, -4}

Resposta de: superlucas007

Explicação passo-a-passo:

(X+2)×x+2 +x =0

x^2 + 4x + 4 +x=0

x^2+5x +4=0

baskara

x=-5 +- raiz (25 - 16)==> 9

x=-5 + 3 /2

x= -1

x=-5-3 /2

x=-4

Conjunto -1 e -4

3x^2 = 2 (x-1 (x-1) +3

3x^2 = 2 (x^2-2x+1)+3

3x^2 = 2x^2 -4x + 5

x^2 +4x -5 =0

x= -4 +- raiz (16 +20) ==36

x = -4 + 6 /2

x = 1

x= -4 - 6 /2

x = -5

Conjunto : 1, -5

Resposta de: Gustavopierro

X1 = 1

X2 = - 5

alterntiva correcta:

Explicação passo-a-passo:

ola.. vamos a isso..

3x² = 2(x-1)² +3 ?

(x-1)² ==== caso notavel quadrado de diferenca:

=> 3x² = 2(x²-2x+1)+3

= > 3x² = 2x²-4x+2+3

=> 3x²- 2x² + 4x-5 =0

=> x²+4x- 5=0

a = 1

b = 4

c = -5

Δ = $b^{2}-4ac$ = (4)²-4*1*(- 5) = 36

x = -b±√Δ/a2

x = -4±√36/2*1

x = -4±6/2

x1= -4+6/2 = 2/2 = 1

x2 = - 4 - 6/2 = -10/2 = - 5

ESPERO TER AJUDADO

Resposta de: liviassiqueira

$\Delta=5^2-4\cdot1\cdot4=25-16=9$

$x=\dfrac{-5\pm\sqrt{9}}{2}=\dfrac{-5\pm3}{2}$

$x'=\dfrac{-5+3}{2}=-1$

$x"=\dfrac{-5-3}{2}=-4$

$S=\{-4,-1\}$

b) 3x=2(x-1)^2+3

$\Delta=(-7)^2-4\cdot2\cdot5=49-40=9$

$x=\dfrac{-(-7)\pm\sqrt{9}}{2\cdot2}=\dfrac{7\pm3}{4}$

$x'=\dfrac{7+3}{4}=\dfrac{10}{4}=\dfrac{5}{2}$

$x"=\dfrac{7-3}{4}=1$

$S=\{1,\frac{5}{2}\}$ .

c) x(x+11)+2(x+21)=0

$\Delta=13^2-4\cdot1\cdot42=169-168=1$

$x=\dfrac{-13\pm1}{2}$

$x'=\dfrac{-13+1}{2}=-6$

$x"=\dfrac{-13-1}{2}=-7$

$S=\{-7,-6\}$

d) 6(x^2-1)-14=5x^2+x

$\Delta=(-1)^2-4\cdot1\cdot(-20)=1+80=81$

$x=\dfrac{-(-1)\pm\sqrt{81}}{2}=\dfrac{1\pm9}{2}$

$x'=\dfrac{1+9}{2}=5$

$x"=\dfrac{1-9}{2}=-4$

$S=\{-4,5\}$

Resposta de: Busca

2.5².5-4.3.1

2.25.5-12

50.5-12

250-12

238

Resposta de: Busca

Ésó dividir um pelo outro:
a) 32/15= 2.1333
b) 5/6= 0.8333
c) 1/4= 0.25
d) 2/5= 0.4

espero ter : )