qualo valor numerico na expressao 4.x+18-5x,para x=12
Soluções para a tarefa
Explicação passo-a-passo:
1)
pelo complemento de quadrados
\begin{lgathered}\mathsf{{x}^{2}-10x+21=0}\\\mathsf{{x}^{2}-10x=-21}\\\mathsf{{x}^{2}-10x+25=25-21}\\\mathsf{{(x-5)}^{2}=4}\\\mathsf{x-5=\pm\sqrt{4}}\\\mathsf{x-5=\pm2}\end{lgathered}
x
2
−10x+21=0
x
2
−10x=−21
x
2
−10x+25=25−21
(x−5)
2
=4
x−5=±
4
x−5=±2
\begin{lgathered}\mathsf{x-5=2\to~x=5+2=7}\\\mathsf{x-5=-2\to~x=5-2=3}\end{lgathered}
x−5=2→ x=5+2=7
x−5=−2→ x=5−2=3
\boxed{\boxed{\mathsf{s=\{3,7\}}}}
s={3,7}
Pela "fórmula de Bháskara":
\mathsf{{x}^{2}-10x+21=0}x
2
−10x+21=0
\begin{lgathered}\mathsf{\Delta={b}^{2}-4ac}\\\mathsf{\Delta={(-10)}^{2}-4.1.21}\\\mathsf{\Delta=100-84=16}\end{lgathered}
Δ=b
2
−4ac
Δ=(−10)
2
−4.1.21
Δ=100−84=16
\begin{lgathered}\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-(-10)\pm\sqrt{16}}{2.1}}\\\mathsf{x=\dfrac{10\pm4}{2}}\end{lgathered}
x=
2a
−b±
Δ
x=
2.1
−(−10)±
16
x=
2
10±4
\begin{lgathered}\mathsf{x_{1}=\dfrac{10+4}{2}=\dfrac{14}{2}=7}\\\mathsf{x_{2}=\dfrac{10-4}{2}=\dfrac{6}{2}=3}\end{lgathered}
x
1
=
2
10+4
=
2
14
=7
x
2
=
2
10−4
=
2
6
=3
\boxed{\boxed{\mathsf{s=\{3,7\}}}}
s={3,7}
2)
Pelo complemento de quadrados
\begin{lgathered}\mathsf{-3{x}^{2}-5x+2=0\div(-3)}\\\mathsf{{x}^{2}+\dfrac{5}{3}x-\dfrac{2}{3}=0}\\\mathsf{{x}^{2}+\dfrac{5}{3}x=\dfrac{2}{3}}\\\mathsf{{x}^{2}+\dfrac{5}{3}x+\dfrac{25}{36}=\dfrac{25}{36}+\dfrac{2}{3}\times(36)}\end{lgathered}
−3x
2
−5x+2=0÷(−3)
x
2
+
3
5
x−
3
2
=0
x
2
+
3
5
x=
3
2
x
2
+
3
5
x+
36
25
=
36
25
+
3
2
×(36)
\begin{lgathered}\mathsf{36{x}^{2}+60x+25=25+24}\\\mathsf{{(6x+5)}^{2}=49}\\\mathsf{6x+5=\pm\sqrt{49}}\\\mathsf{6x+5=\pm7}\end{lgathered}
36x
2
+60x+25=25+24
(6x+5)
2
=49
6x+5=±
49
6x+5=±7
\begin{lgathered}\mathsf{6x+5=7}\\\mathsf{6x=7-5}\\\mathsf{6x=2\to~x=\dfrac{2\div2}{6\div2}=\dfrac{1}{3}}\end{lgathered}
6x+5=7
6x=7−5
6x=2→ x=
6÷2
2÷2
=
3
1
\begin{lgathered}\mathsf{6x+5=-7}\\\mathsf{6x=-5-7}\\\mathsf{6x=-12\to~x=-\dfrac{12}{6}=-2}\end{lgathered}
6x+5=−7
6x=−5−7
6x=−12→ x=−
6
12
=−2
\boxed{\boxed{\mathsf{s=\{-2,\dfrac{1}{3}\}}}}
s={−2,
3
1
}
Pela "fórmula de Bháskara":
\begin{lgathered}\mathsf{-3{x}^{2}-5x+2=0\times(-1)}\\\mathsf{3{x}^{2}+5x-2=0}\end{lgathered}
−3x
2
−5x+2=0×(−1)
3x
2
+5x−2=0
\begin{lgathered}\mathsf{\Delta={b}^{2}-4ac}\\\mathsf{\Delta={5}^{2}-4.3.(-2)}\\\mathsf{\Delta=25+24=49}\end{lgathered}
Δ=b
2
−4ac
Δ=5
2
−4.3.(−2)
Δ=25+24=49
\begin{lgathered}\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-5\pm\sqrt{49}}{2.3}}\\\mathsf{x=\dfrac{-5\pm7}{6}}\end{lgathered}
x=
2a
−b±
Δ
x=
2.3
−5±
49
x=
6
−5±7
\mathsf{x_{1}=\dfrac{-5-7}{6}=-\dfrac{12}{6}=-2}x
1
=
6
−5−7
=−
6
12
=−2
\mathsf{x_{2}=\dfrac{-5+7}{6}=\dfrac{2}{6}=\dfrac{1}{3}}x
2
=
6
−5+7
=
6
2
=
3
1
\boxed{\boxed{\mathsf{s=\{-2,\dfrac{1}{3}\}}}}
s={−2,
3
1
}