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a) Tìm h(x) = f(x) - g(x)
f(x) - g(x) = (-2x2 - 3x3 - 5x + 5x3 - x + x2 + 4x + 3 + 4x2) - (2x2 - x3 + 3x + 3x3 + x2 - x - 9x + 2)
= -2x2 - 3x3 - 5x + 5x3 - x + x2 + 4x + 3 + 4x2 - 2x2 + x3 - 3x - 3x3 - x2 + x + 9x - 2
= (-2x2 + x2 + 4x2 - 2x2 - x2) + (-3x3 + 5x3 + x3 - 3x3) + (-5x - x + 4x - 3x + x + 9x) + (3 - 2)
= 5x + 1
Vậy h(x) = 5x + 1
b) Tìm nghiệm của đa thức h(x)
Cho h(x) = 0
\(\Leftrightarrow\) 5x + 1 = 0
5x = 0 + 1
5x = 1
x = \(\dfrac{1}{5}\)
Vậy x = \(\dfrac{1}{5}\) là nghiệm của đa thức h(x).
a: \(M\left(x\right)=2x^2+3\)
\(N\left(x\right)=3x^3-2x^2+x\)
b: \(M\left(x\right)+N\left(x\right)=3x^3+x+3\)
\(M\left(x\right)-N\left(x\right)=2x^2+3-3x^3+2x^2-x=-3x^3+2x^2-x+3\)
a) M(x) = 3x3 – 5x + 2x2 + 9 = 3x3 + 2x2 - 5x + 9
N(x) = 5 – 4x + 3x3 – 2x2 = 3x3 - 2x2 - 4x + 5
b) M (x) + N(x) = 3x3 + 2x2 - 5x + 9 + 3x3 - 2x2 - 4x + 5
= ( 3x3 + 3x3 ) + ( 2x2 - 2x2 ) + ( -5x - 4x) + ( 9 + 5 )
= 6x3 - 9x + 14
M (x) - N (x) = 3x3 + 2x2 - 5x + 9 - (3x3 - 2x2 - 4x + 5)
= 3x3 + 2x2 - 5x + 9 - 3x3 + 2x2 + 4x - 5
= ( 3x3 - 3x3 ) + ( 2x2 + 2x2 ) + ( -5x + 4x ) + ( 9 - 5)
= 4x2 - x + 4
;y=-1
; Q(x) = x2 – 9.; R(x) = 3x2 – 4x
)
21:
a: \(f\left(x\right)=4x^4-x^3-4x^2+x-1\)
\(g\left(x\right)=x^4+4x^3+x-5\)
b: f(x)-g(x)
=4x^4-x^3-4x^2+x-1-x^4-4x^3-x+5
=3x^4-5x^3-4x^2+4
f(x)+g(x)
=4x^4-x^3-4x^2+x-1+x^4+4x^3+x-5
=5x^4+3x^3-4x^2+2x-6
c: g(-1)=1-4-1-5=-9
a) Ta có: \(M\left(x\right)=3x^3+x^2+4x^4-x-3x^3+5x^4+2x^2-6\)
\(=\left(4x^4+5x^4\right)+\left(3x^3-3x^3\right)+\left(x^2+2x^2\right)-x-6\)
\(=9x^4+3x^2-x-6\)
Ta có: \(N\left(x\right)=-2x^2-x^4+4x^3-x^2-5x^3+3x+5+x\)
\(=-x^4+\left(4x^3-5x^3\right)+\left(-2x^2-x^2\right)+\left(3x+x\right)+5\)
\(=-x^4-x^3-3x^2+4x+5\)
c) Ta có: M(x)+N(x)
\(=9x^4+3x^2-x-6-x^4-x^3-3x^2+4x+5\)
\(=8x^4-x^3+3x-1\)
a)h(x)=f(x)-g(x)
=(2x3 +3x2 -2x +3)-(2x3 +3x2 -7x +2)
=2x3 + 3x2 - 2x +3 - 2x3 -3x2 + 7x -2
=5x+1
b)h(x)=5x+1=0
=>5x=-1
x=\(\frac{-1}{5}\)
a: F(x)=3x^3-2x^2+5x-7
G(x)=3x^3-2x^2+5x+7x^2+3=3x^3+5x^2+5x+3
Bậc của F(x),G(x) đều là 3
b: N(x)=G(x)-F(x)
\(=3x^3+5x^2+5x+3-3x^3+2x^2-5x+7=7x^2+10\)
M(x)=2F(x)+G(x)
\(=6x^3-4x^2+10x-14+3x^3+5x^2+5x+3\)
\(=9x^3+x^2+15x-11\)
c: x^2-3x=0
=>x=0 hoặc x=3
\(M\left(0\right)=9\cdot0^3+0^2+15\cdot0-11=-11\)
\(M\left(3\right)=9\cdot3^3+3^2+15\cdot3-11=286\)
d: N(x)=7x^2+10>=10
Dấu = xảy ra khi x=0
a) M+N+P= 3x3-4x2-x+5+x3-2x2+5x-4+3x3-x2-5x+10
=3x3+x3+3x3-4x2-2x2-x2-x+5x-5x+5-4+10
=7x3-7x2-x+11
b)M-N-P=3x3-4x2-x+5-(x3-2x2+5x-4)-(3x3-x2-5x+10)
=3x3-4x2-x+5-x3+2x2-5x+4-3x3+x2+5x-10
=3x3-x3-3x3-4x2+2x2+x2-x-5x+5x+5+4-10
=-x3-x2-x-1
c)N-M-P=x3-2x2+5x-4-(3x3-4x2-x+5)-(3x3-x2-5x+10)
=x3-2x2+5x-4-3x3+4x2+x-5-3x3+x2+5x-10
=x3-3x3-3x3-2x2+4x2+x2+5x+x+5x-4-5-10
=-5x3+3x2+11x-19
d)P-N-M=3x3-x2-5x+10-(x3-2x2+5x-4)-(3x3-4x2-x+5)
=3x3-x2-5x+10-x3+2x2-5x+4-3x3+4x2+x-5
=3x3-x3-3x3-x2+2x2+4x2-5x-5x+x+10+4-5
=-x3+5x2-9x+9