Toán Lớp 9: tìm nghiệm các phương trình sau:
a, 2x^2 – 3x -5 = 0
b, x^2 – 6x + 8 = 0
c, 9x^2 – 12x + 4 = 0
d, -3x^2 + 4x – 4 = 0
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Toán Lớp 9: tìm nghiệm các phương trình sau:
a, 2x^2 – 3x -5 = 0
b, x^2 – 6x + 8 = 0
c, 9x^2 – 12x + 4 = 0
d, -3x^2 + 4x – 4 = 0
Comments ( 2 )
$ a) 2x^2 – 3x -5 = 0 $
$ ⇒ 2x(x+1)-5(x+1)=0 $
$ ⇒ ( 2x-5)(x+1)=0 $
$ ⇒ $ \(\left[ \begin{array}{l}2x-5=0⇒2x=5⇒x=\dfrac{5}{2}\\x+1=0⇒x=-1\end{array} \right.\)
$ b) x^2 – 6x + 8 = 0 $
$ ⇒ x^2 – 2x – 4x + 8 = 0 $
$ ⇒ x(x-2) – 4(x-2) = 0 $
$ ⇒ (x-2)(x-4)= 0 $
$ ⇒ $ \(\left[ \begin{array}{l}x-2=0⇒x=2\\x-4=0⇒x=4\end{array} \right.\)
$ c) 9x^2 – 12x + 4 = 0 $
$ ⇒ (3x)^2-2.3x.2+2^2=0 $
$ ⇒ (3x-2)^2=0 $
$ ⇒ 3x-2=0 $
$ ⇒ 3x=2 $
$ ⇒ x = \dfrac{2}{3} $
$a)$ $2x^2\:-\:3x\:-5\:=\:0$
⇔2x(x+1)−5(x+1)=0
⇔(x+1)(2x−5)=0
\(⇔\left[ \begin{array}{l}x+1=0\\2x-5=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-1\\x=\dfrac{5}{2}\end{array} \right.\)
$\text{Vậy pt có nghiệm S = {-1 ; $\dfrac{5}{2}$}}$
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$b)$ $x^2-6x+8=0$
$⇔\left(x−4\right)\left(x−2\right)=0$
\(⇔\left[ \begin{array}{l}x-4=0\\\:x-2=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\)
$\text{Vậy pt có nghiệm S = {4 ; 2}}$
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$c)$ $9x^2 – 12x + 4 = 0$
⇔(3x) ^2 −2(3x)(2)+2 ^2 =0
⇔(3x−2) ^2 =0
⇔3x−2=0
⇔3x=2
$⇔x=\dfrac{2}{3}$
$\text{Vậy pt có nghiệm S = {$\dfrac{2}{3}$}}$
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$d)$ $-3x^2\:+\:4x\:-\:4\:=\:0$
⇔−3x^2+4x+−4=0
$⇔x=\dfrac{-\left(4\right)\pm \sqrt{\left(4\right)^2-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$
$⇔x=\dfrac{-\left(4\right)\pm \sqrt{32}}{-6}$
$\text{Vậy pt vô nghiệm}$