Giải phương trình : (3x-1)(2x-3)(2x-3)(x+5)=0 3x-15 = 2x(x-5) 31/08/2021 Bởi Harper Giải phương trình : (3x-1)(2x-3)(2x-3)(x+5)=0 3x-15 = 2x(x-5)
`(3x-1)(2x-3)(2x-3)(x+5)=0` `⇔(3x-1)(2x-3)^2(x+5)=0` TH1: `3x – 1 = 0` `⇔ x = 1/3` TH2: `(2x-3)^2=0` `⇔2x-3=0` `⇔x=3/2` TH3: `x+5=0` `⇔x=-5` `3x-15=2x(x-5)` `⇔3x-15=2x^2-10x` `⇔-2x^2+13x-15=0` `⇔-2x^2+10x+3x-15=0` `⇔-2x(x-5)+3(x-5)=0` `⇔(3-2x)(x+5)=0` TH1: `3-2x=0` `⇔x=3/2` TH2: `x+5=0` `⇔x = -5` Bình luận
`a,` `(3x-1)(2x-3)(2x-3)(x+5)=0` \(⇒\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.⇒\left[ \begin{array}{l}x=\dfrac13\\x=\dfrac32\\x=-5\end{array} \right.\) `Vậy` `S={1/3;3/2;-5}` `b,` `3x-15=2x(x-5)` `⇒3(x-5)=2x(x-5)` \(⇒\left[ \begin{array}{l}3=2x\\x-5\end{array} \right.⇒\left[ \begin{array}{l}x=\dfrac32\\x=5\end{array} \right.\) `Vậy` `S={3/2;5}` Bình luận
`(3x-1)(2x-3)(2x-3)(x+5)=0`
`⇔(3x-1)(2x-3)^2(x+5)=0`
TH1: `3x – 1 = 0`
`⇔ x = 1/3`
TH2: `(2x-3)^2=0`
`⇔2x-3=0`
`⇔x=3/2`
TH3: `x+5=0`
`⇔x=-5`
`3x-15=2x(x-5)`
`⇔3x-15=2x^2-10x`
`⇔-2x^2+13x-15=0`
`⇔-2x^2+10x+3x-15=0`
`⇔-2x(x-5)+3(x-5)=0`
`⇔(3-2x)(x+5)=0`
TH1: `3-2x=0`
`⇔x=3/2`
TH2: `x+5=0`
`⇔x = -5`
`a,` `(3x-1)(2x-3)(2x-3)(x+5)=0`
\(⇒\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.⇒\left[ \begin{array}{l}x=\dfrac13\\x=\dfrac32\\x=-5\end{array} \right.\)
`Vậy` `S={1/3;3/2;-5}`
`b,` `3x-15=2x(x-5)`
`⇒3(x-5)=2x(x-5)`
\(⇒\left[ \begin{array}{l}3=2x\\x-5\end{array} \right.⇒\left[ \begin{array}{l}x=\dfrac32\\x=5\end{array} \right.\)
`Vậy` `S={3/2;5}`