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a. M(x) + N(x) = 3x3 - 3x + x2 + 5 + 2x2 - x + 3x3 + 9
= (3x3 + 3x3) + ( x2 + 2x2 ) + ( -3x - x ) + (5 + 9)
= 6x3 + 3x2 - 4x + 14
b. M(x) + N(x) - P(x) = 6x3 + 3x2 + 2x
=> 6x3 + 3x2 - 4x + 14 - P(x) = 6x3 + 3x2 + 2x
=> 6x3 + 3x2 - 4x + 14 - ( 6x3 + 3x2 + 2x) = P(x)
=> 6x3 + 3x2 - 4x + 14 - 6x3 - 3x2 - 2x = P(x)
=> (6x3 - 6x3 ) + (3x2 - 3x2 ) + (-4x - 2x ) + 14 = P(x)
=> -6x + 14 = P(x)
Ta có : -6x + 14 = 0
=> -6x = -14
=> x = 7/3
=> Đa thức P(x) = -6x + 14 có nghiệm là 7/3
=>
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Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
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a) \(f\left(-2\right)=2.\left(-2\right)^2+3.\left(-2\right)-5\)
\(f\left(-2\right)=8-6-5=-3\)
b) \(f\left(x\right)+g\left(x\right)\)
\(=2x^2+3x-5+3x^2+2x-7\)
\(=5x^2+5x-12\)
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P(x)+Q(x)
=3x^2y-2x+5xy^2-7y^2+3xy^2-7y^2-9x^2y-x-5
=8xy^2-14y^2-6x^2y-3x-5
=>Chọn A
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a/ M(x)+N(x)=(3x3+3x3)+(x2+2x2)-(3x+x)+(5+9)
=6x3+3x2-4x+14
b/ Ta có: M(x)+N(x)-P(x)=6x3+3x2+2x
=> P(x)=M(x)+N(x)-6x3+3x2+2x=-6x
c/ P(x)=-6x=0
=> x=0 là nghiệm đa thức P(x)
d/ Ta có: x2+4x+5
=x.x+2x+2x+2.2+1
=x(x+2)+2(x+2)+1
=(x+2)(x+2)+1
=(x+2)2+1
Mà (x+2)2\(\ne0\)=> Đa thức trên \(\ge1\)
=> Đa thức trên vô nghiệm.
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f(x)+g(x)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3-x^4-x^2+x^3-x+5-5x^3+x^2+3x^4
=(-3x^2-x^2-x^2+x^2)+(x-x)-(1-5)+(x^4+3x^4-x^4+3x^4)-(x^3-2x^3-x^3+5x^3)
=-4x^2+4+6x^4+3x^3