1. a, f(x)+(3x ²-5x)=2x ²-4x.
b, (2x ²-3x-4)-f(x)=3x ²+4x-5.
2. Cho đa thức P(x)+Q(x)=5x ²-4x+1 và P(x)-Q(x)=x ²+2x-5. Tìm đa thức P(x) và Q(x).
3. Cho đa thức f(x)=ax ²+bx +c, biết 13a+b+2c=0. Chứng tỏ rằng: f(-2).f(3) ≤0.
1. a, f(x)+(3x ²-5x)=2x ²-4x. b, (2x ²-3x-4)-f(x)=3x ²+4x-5. 2. Cho đa thức P(x)+Q(x)=5x ²-4x+1 và P(x)-Q(x)=x ²+2x-5. Tìm đa thức P(x) và Q(x). 3. Ch
By Audrey
`1)`
`a) f(x) + (3x^2 – 5x) = 2x^2 – 4x`
`=> f(x) = 2x^2 – 4x – (3x^2 – 5x)`
`=> f(x) = 2x^2 – 4x – 3x^2 + 5x`
`=> f(x) = (2x^2 – 3x^2) + (-4x + 5x)`
`=> f(x) = -x^2 + x`
`b) (2x^2 – 3x – 4) – f(x) = 3x^2 + 4x – 5`
`=> (2x^2 – 3x – 4) – (3x^2 + 4x – 5)= f(x)`
`=> 2x^2 – 3x – 4 – 3x^2 – 4x + 5 = f(x)`
`=> (2x^2 – 3x^2) – (3x + 4x) + (-4 + 5) = f(x)`
`=> -x^2 – 7x + 1 = f(x)`
`2)`
`P(x) + Q(x) = 5x^2 – 4x + 1`
`P(x) – Q(x) = x^2 + 2x – 5`
`=> P(x) = {[P(x) + Q(x)] + [P(x) – Q(x)]} : 2 = (5x^2 – 4x + 1 + x^2 + 2x – 5) : 2`
`= (6x^2 – 2x – 4) : 2`
`= 3x^2 – x – 2`
`=> Q(x) = [P(x) + Q(x)] – P(x) = (5x^2 – 4x + 1) – (3x^2 – x – 2)`
`= 5x^2 – 4x + 1 – 3x^2 + x + 2`
`= 2x^2 – 3x + 3`
Vậy `P(x) = 3x^2 – x – 2; Q(x) = 2x^2 – 3x + 3`
`3)`
`f(x) = ax^2 + bx + c`
`+) => f(-2) = a. (-2)^2 + b. (-2) + c`
`= 4a – 2b + c`
`+) => f(3) = a. 3^2 + b. 3 + c`
`= 9a + 3b + c`
`= 13a – 4a + b + 2b + 2c – c`
`= (13a + b + 2c) – (4a – 2b + c)`
`= – (4a – 2b + c)`
`= – f(-2)`
`=> f(3) = – f(-2)`
`=> f(-2). f(3) = f(-2). -f(-2) = – [f(-2)]^2 <= 0`
`=> f(-2). f(3) <= 0`
`=> đpcm`
a) `f(x) + (3x^2 – 5x) = 2x^2 -4x`
`=> f(x)= (2x^2 – 4x) – (3x^2 -5x)`
`=> f(x) = 2x^2 – 4x – 3x^2 + 5x`
`=> f(x)= (2x^2 – 3x^2) – (4x-5x)`
`=> f(x)= -x^2 + x`
Vậy `f(x) = -x^2 +x`
b) `(2x^2 – 3x -4) – f(x) = 3x^2 + 4x-5`
`=> f(x) = (2x^2 – 3x -4) – (3x^2 + 4x – 5)`
`=> f(x) = 2x^2 – 3x -4 – 3x^2 – 4x +5`
`=>f(x) = (2x^2 – 3x^2)- (3x + 4x) + (5-4)`
`=> f(x)= -x^2 – 7x + 1`
Vậy `f(x) = -x^2 – 7x +1`
2)
`P(x) = [( 5x^2 – 4x + 1) + (x^2 + 2x -5)]:2`
`= (5x^2 – 4x +1 + x^2 + 2x -5):2`
`=[ (5x^2 +x^2) – (4x – 2x) + (1-5)] :2`
`= (6x^2 – 2x – 4):2`
`= 3x^2 – x -2`
`Q(x) = (5x^2 – 4x +1 ) – ( 3x^2 – x -2)`
`= 5x^2 – 4x +1 – 3x^2 + x + 2`
`= (5x^2 – 3x^2) – (4x -x) + (1+2)`
`= 2x^2 – 3x + 3`
Vậy `P(x) = 3x^2 – x -2` ; `Q(x) = = 2x^2 – 3x + 3`
3) Cho `x= -2`
`=> f(-2) = a. (-2)^2 + b.(-2) + c`
`=> f(-2) = 4a – 2b +c`
Cho `x= 3`
`=> f(3) = a.3^2 + b.3 + c`
`=> f(3) = 9a + 3b+c`
Do đó: `f(-2) + f(3) = 4a – 2b + c + 9a + 3b +c`
`f(-2) + f(3) = 13a + b + 2c =0`
`=> f(-2) = -f(3)`
`=> f(-2) . f(3) = -[f(3)]^2 le 0`
Vậy `f(-2) . f(3) le 0`