Bài `4:`
`a)`
`x^3 + 4x = 0`
`=> x (x^2 + 4) = 0`
`=> x = 0` hoặc `x^2 + 4=0`
`+) x = 0`
`+) x^2 + 4 = 0`
`=> x^2 = -4` (không xảy ra do `x^2 \ge 0 \forall x`)
Vậy `x=0`
`b)`
`x (x-2) + 3 (x-2) =0`
`=> (x+3)(x-2)=0`
`=>x+3=0` hoặc `x-2=0`
`+) x+ 3 = 0 => x = 0 - 3 => x = -3`
`+) x - 3 =0 => x = 0 +2 =>x=2`
Vậy `x \in {-3 ; 2}`
`c)`
`3x (2x-1) - 2x + 1 =0`
`=> 3x (2x-1) -(2x-1)=0`
`=> (3x-1)(2x-1)=0`
`=>3x-1=0` hoặc `2x-1=0`
`+) 3x - 1= 0`
`=> 3x = 1`
`=> x = 1/3`
`+) 2x - 1 = 0`
`=> 2x = 1`
`=> x = 1/2`
Vậy `x \in {1/3 ; 1/2}`
`d)`
`(2x+1)^2 - (3x+2)^2=0`
`=> [ (2x+1) + (3x+2)] . [ (2x+1) - (3x+2)] =0`
`=> (2x + 1 + 3x + 2)(2x+ 1 -3x-2)=0`
`=> (5x + 3)(-x-1)=0`
`=>5x+3=0` hoặc `-x-1=0`
`+) 5x + 3=0`
`=> 5x =-3`
`=>x=-3/5`
`+) -x - 1= 0`
`=> -x = 1`
`=> x = -1`
Vậy `x \in {-3/5 ; -1}`
`e)`
`x^2 - 3x = -2`
`=> x^2 - 3x+ 2=0`
`=> x^2 - 2x - x + 2=0`
`=> x (x-2) - (x-2)=0`
`=> (x-1)(x-2)=0`
`=>x-1=0` hoặc `x-2=0`
`+) x -1 = 0 => x = 0 +1 =>x=1`
`+) x-2=0=>x=0+2 =>x=2`
Vậy `x \in {1;2}`
