• Matéria: Matemática
  • Autor: SofiaGabriele
  • Perguntado 9 anos atrás

ME AJUDEM PELO AMOR DE DEUS!
a) X²+7x=0
b) -3x/2+9x=0
c)2x²+3x=0
d) (x+5)(x-1)=2x
e) (x+5)²=2x+23
f) (x-3)²=9
g (x+2)²=4

Respostas

respondido por: korvo
0
Olá Sofia,

dadas a equações do 2º grau, podemos resolver da seguinte forma:

 x^{2} +7x=0\\
x(x+7)=0\\\\
\begin{cases}x'=0\end{cases}\begin{cases}x+7=0\\
x''=-7\end{cases}\\\\\\
\boxed{S=\{0,-7\}}

____________________

- \dfrac{3x^2}{2}+9x=0\\\\
 x\left(- \dfrac{3}{2}x+9\right)=0\\\\\\
\begin{cases}x'=0\end{cases}\begin{cases}- \dfrac{3}{2}x+9=0\\\\
- \dfrac{3}{2}x=-9\\\\
- 3x=2*(-9)\\
-3x=-18\\\\
x= \dfrac{-18}{-3}\\\\
x''=6   \end{cases}\\\\\\
\boxed{S=\{0,6\}}

____________________

2 x^{2} +3x=0\\
x(2x+3)=0\\\\
\begin{cases}x'=0\end{cases}\begin{cases}2x+3=0\\
2x=-3\\
x''=- \dfrac{3}{2} \end{cases}\\\\\\
\boxed{S=\left\{0,- \dfrac{3}{2}\right\}}

____________________

(x+5)(x-1)=2x\\
 x^{2} -x+5x-5=2x\\
 x^{2} +4x-2x-5=0\\
 x^{2} +2x-5=0\\\\
\Delta=b^2-4ac\\
\Delta=2^2-4*1*(-5)\\
\Delta=4+20\\
\Delta=24\\\\
x= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-2\pm \sqrt{24} }{2*1}= -1\pm2 \sqrt{6}\\\\\\
\boxed{S=\{-1+2 \sqrt{6},~-1-2 \sqrt{6}\}}

____________________

(x+5)^2=2x+23\\
(x+5)(x+5)=2x+23\\
 x^{2} +5x+5x+25=2x+23\\
 x^{2} +10x+25-2x-23=0\\
 x^{2} +8x+2=0\\\\
\Delta=8^2-4*1*2\\
\Delta=64-8\\
\Delta=56\\\\\\
x= \dfrac{-8\pm \sqrt{56} }{2*1}=-4\pm2 \sqrt{14}\\\\\\
\boxed{S=\{-4+2 \sqrt{14},~-4-2 \sqrt{14}}

____________________

(x-3)^2=9\\
(x-3)=\pm \sqrt{9}\\\\
(x-3)=\pm3~\to~\begin{cases}x-3=3~\to~x=3+3~\to~x'=6\\
x-3=-3~\to~x=-3+3~\to~x''=0\end{cases}\\\\\\
\boxed{S=\{0,6\}}

____________________

(x+2)^2=4\\
(x+2)=\pm \sqrt{4}\\\\
x+2=\pm2~\to~\begin{cases}x+2=2~\to~x=2-2~\to~x'=0\\
x+2=-2~\to~x=-2-2~\to~x''=-4 \end{cases}\\\\\\
\boxed{S=\{0,-4\}}

Espero ter ajudado e tenha ótimos estudos =))
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