Bài toán 3 : Tìm x biết.
a) 4x(x + 1) = 8(x + 1) g) 5x(x – 2000) – x + 2000 = 0
b) x(x – 1) – 2(1 – x) = 0 h) x – 4x = 0
c) 2x(x – 2) – (2 – x) = 0 k) (1 – x) – 1 + x = 0
d) (x – 3) + 3 – x = 0 m) x + 6x = 0
Bài toán 3 : Tìm x biết.
a) 4x(x + 1) = 8(x + 1) g) 5x(x – 2000) – x + 2000 = 0
b) x(x – 1) – 2(1 – x) = 0 h) x – 4x = 0
c) 2x(x – 2) – (2 – x) = 0 k) (1 – x) – 1 + x = 0
d) (x – 3) + 3 – x = 0 m) x + 6x = 0
`a, 4x(x+1)=8(x+1)`
`⇔4x(x+1)-8(x+1)=0`
`⇔(x+1)(4x-8)=0`
`⇔` \(\left[ \begin{array}{l}x+1=0\\4x-8=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\)
`b, x(x-1)-2(1-x)=0`
`⇔ x(x-1)+2(x-1)=0`
`⇔ (x+2)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x-1=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
`c, 2x(x-2)-(2-x)=0`
`⇔ 2x(x-2)+(x-2)=0`
`⇔ (2x+1)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}2x+1=0\\x-2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=2\end{array} \right.\)
`d, (x-3) +3-x=0`
`⇔x-3 + 3-x=0`
`⇔0=0`
`⇒x∈R`
`g, 5x(x-2000) – x+2000=0`
`⇔5x(x-2000)-(x-2000)=0`
`⇔(5x-1)(x-2000)=0`
`⇔` \(\left[ \begin{array}{l}5x-1=0\\x-2000=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=2000\end{array} \right.\)
`h, x – 4x = 0`
`⇔ -3x = 0`
`⇔ x = 0`
`k, (1-x) – 1+x=0`
`⇔1-x-1+x=0`
`⇔0=0`
`⇔x∈R`
`m, x +6x = 0`
`⇔ 7x =0`
`⇔x=0`
a) 4x(x + 1) = 8(x + 1)
=>4x(x + 1) – 8(x +1)= 0
=> (x+1)(4x-8) =0
=> x+1 = 0 hoặc 4x-8 = 0
=> x= -1 hoặc x= 2
Vậy x ∈ { -1 ; 2}
b) x(x – 1) – 2(1 – x) = 0
=> x(x-1) + 2(x-1) = 0
=> (x-1)(x+2) =0
=> x-1 = 0 hoặc x+2 =0
=> x= 1 hoặc x= -2
Vậy x ∈ { 1 ; -2}
c) 2x(x – 2) – (2 – x) = 0
=> 2x(x-2) + ( x-2) = 0
=> (x-2)(2x+1) = 0
=> x-2 = 0 hoặc 2x+1 = 0
=> x = 2 hoặc x= -1/2
Vậy x { 2 ; -1/2}
d) (x – 3) + 3 – x = 0
=> (x-3) – ( x-3) = 0
=> (x-3)(1-1) = 0
=> ( x-3).0 =0
=> x€R
Vậy x€R
g) 5x(x – 2000) – x + 2000 = 0
=> 5x( x-2000) -(x-2000) =0
=> (x-2000)(5x-1) =0
=> x-2000 = 0 hoặc 5x-1 = 0
=> x= 2000 hoặc x= 1/5
Vậy x∈ { 2000; 1/5}
h) x – 4x = 0
=> -3x=0
=> x= 0
Vậy x=0
k) (1 – x) – 1 + x = 0
=> (1-x) -(1-x) =0
=>(1-x)(1-1) = 0
=> (1-x).0 = 0
=> x € R
Vậy x € R
m) x + 6x = 0
=> 7x =0
=> x=0
Vậy x=0