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\(\text{a)}f\left(x\right)-g\left(x\right)+h\left(x\right)=\left(x^3-2x^2+3x+1\right)-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
\(=x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)
\(=\left(x^3-x^3\right)+\left(-2x^2+2x^2\right)+\left(3x-x\right)+\left(1+1-1\right)\)
\(=2x+1\)
\(\text{b)Vì f(x)-g(x)+h(x)=0}\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x\) \(=0-1=-1\)
\(\Rightarrow\) \(x\) \(=\left(-1\right):2=\dfrac{-1}{2}\)
\(\text{Vậy x=}\dfrac{-1}{2}\text{ thì f(x)-g(x)+h(x)=0}\)
a: \(f\left(x\right)-g\left(x\right)+h\left(x\right)\)
\(=2x^3-2x^2+4x+2x^2-1=2x^3+4x-1\)
b: f(x)-g(x)+h(x)=0
\(\Leftrightarrow2x^3+4x-1=0\)
\(\Leftrightarrow x\simeq0,2428\)
a) Ta có: \(B\left(x\right)=-2x^3+2x^2+12+5x^2-9x\)
\(=-2x^3+7x^2-9x+12\)
b) Ta có: A(x)+B(x)
\(=4x^3-7x^2+3x-12-2x^3+7x^2-9x+12\)
\(=2x^3-6x\)
b) Ta có: A(x)-B(x)
\(=4x^3-7x^2+3x-12+2x^3-7x^2+9x-12\)
\(=6x^3-14x^2+12x-24\)
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2
= x 3 + - 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
P ( x ) − Q ( x ) = − 2 x 3 + x 2 + x − 2 − x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 − x 3 − 3 x − 1 = − 2 x 3 − x 3 + x 2 + ( x − 3 x ) − 2 − 1 = − 3 x 3 + x 2 − 2 x − 3
Chọn đáp án B
a) 2x(x+3) – 3x2(x+2) + x(3x2 + 4x – 6)
= (2x . x + 2x . 3) – (3x2 . x + 3x2 . 2) + (x . 3x2 + x . 4x – x . 6)
= 2x2 + 6x – (3x3 + 6x2) + (3x3 + 4x2 - 6x)
= 2x2 + 6x – 3x3 – 6x2 + 3x3 + 4x2 - 6x
= (– 3x3 + 3x3 ) + (2x2 - 6x2 + 4x2 ) + (6x – 6x)
= 0 + 0 + 0
= 0
b) 3x(2x2 – x) – 2x2(3x+1) + 5(x2 – 1)
= [3x . 2x2 + 3x . (-x)] – (2x2 . 3x + 2x2 . 1) + [5x2 + 5 . (-1)]
= 6x3 – 3x2 – (6x3 +2x2) + 5x2 – 5
= 6x3 – 3x2 – 6x3 - 2x2 + 5x2 – 5
= (6x3 – 6x3 ) + (-3x2 – 2x2 + 5x2) – 5
= 0 + 0 – 5
= - 5
`3x(2x^2-3x+1)`
`= 3x*2x^2+3x*(-3x)+3x*1`
`= 6x^3-9x^2+3x`
`3x (2x^2 - 3x + 1)`
`= 3x . 2x^2 - 3x . 3x + 3x . 1`
`= 6x^3 - 9x^2 + 3x`
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2 = x 3 + − 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
M ( x ) = P ( x ) + Q ( x ) = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 3 + x 2 + ( x + 3 x ) − 2 + 1 = − x 3 + x 2 + 4 x − 1
Bậc của M ( x ) = - x 3 + x 2 + 4 x - 1 l à 3
Chọn đáp án C
\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)
\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)
\(=2\cdot\left(-1\right)^2-3\cdot\left(-1\right)+1=0\)
\(2x^2-3x+1=2x^2-2x-x+1=\left(2x-1\right)\left(x+1\right)\)
Tại x=-1 , giá trị biểu thức là :
\(\left[2\cdot\left(-1\right)-1\right]\left(-1+1\right)=0\)
Ta có: (-2x2) . (3x – 4x3 + 7 – x2)
= (-2x2) . 3x + (-2x2) . (-4x3) + (-2x2) . 7 + (-2x2) . (-x2)
= [(-2).3] . (x2 . x) + [(-2).(-4)] . (x3 . x2) + [(-2).7] . x2 + [(-2).(-1)] . (x2 . x2)
= -6x3 + 8x5 + (-14)x2 + 2x4
= 8x5 +2x4 -6x3 – 14x2
2x2 -3x+ 12=0
delta<0 => PT vô nghiệm