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f(x) = anxn + an – 1xn– 1 + … + a1x + ao
-
g(x) = bnxn + bn – 1xn– 1 + … + b1x + bo
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f(x) - g(x) = (an - bn)xn + (an– 1 - bn – 1)xn– 1 + ..… + (a1 - b1)x + (ao - bo)
`a)f(x)-g(x)`
`=x^3-2x^2+3x+1-(x^3+x-1)`
`=x^3-2x^2+3x+1-x^3-x+1`
`=(x^3-x^3)+(3x-x)-2x^2+2`
`=-2x^2+2x+2=0`
`b)f(x)-g(x)+h(x)=0`
`<=>-2x^2+2x+2+2x^2-1=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
Vậy `x=-1/2` thì `f(x)-g(x)+h(x)=0`
a) \(f\left(x\right)-g\left(x\right)\) hay \(x^3-2x^2+3x+1-x^3-x+1=-2x^2+2x+2\)
b) \(f\left(x\right)-g\left(x\right)+h\left(x\right)=0\) hay \(-2x^2+2x+2+2x^2-1=2x+1\Rightarrow2x+1=0\Rightarrow x=-\dfrac{1}{2}\)
`a,`
`F(x)=4x^4-2+2x^3+2x^4-5x+4x^3-9`
`F(x)=(2x^4+4x^4)+(2x^3+4x^3)-5x+(-2-9)`
`F(x)=6x^4+6x^3-5x-11`
`b,`
`K(x)=F(x)+G(x)`
`K(x)=(6x^4+6x^3-5x-11)+(6x^4+6x^3-x^2-5x-27)`
`K(x)=6x^4+6x^3-5x-11+6x^4+6x^3-x^2-5x-27`
`K(x)=(6x^4+6x^4)+(6x^3+6x^3)-x^2+(-5x-5x)+(-11-27)`
`K(x)=12x^4+12x^3-x^2-10x-38`
`c,`
`H(x)=F(x)-G(x)`
`H(x)=(6x^4+6x^3-5x-11)-(6x^4+6x^3-x^2-5x-27)`
`H(x)=6x^4+6x^3-5x-11-6x^4-6x^3+x^2+5x+27`
`H(x)=(6x^4-6x^4)+(6x^3-6x^3)+x^2+(-5x+5x)+(-11+27)`
`H(x)=x^2+16`
Đặt `x^2+16=0`
Ta có: \(x^2\ge0\text{ }\forall\text{ }x\)
`->`\(x^2+16\ge16>0\text{ }\forall\text{ }x\)
`->` Đa thức `H(x)` vô nghiệm.
Ta có: f(x) - g(x) = x3 - 2x2 + 3x + 1 - (x3 + x - 1) = -2x2 + 2x
f(x) - g(x) + h(x) = -2x2 + 2x + 2x2 - 1 = 2x - 1
Mà: f(x) - g(x) + h(x) = 0
⇒ 2x - 1 = 0
\(\Leftrightarrow x=\dfrac{1}{2}\)
a) Ta có:+) f(x) = 2x2(x - 1) - 5(x - 2) - 2x(x - 2)
f(x) = 2x3 - 2x2 - 5x + 10 - 2x2 + 2x
f(x) = 2x3 - 4x2 - 3x + 10
f(x) = 2x3 - 2x2 - 5x + 10
+) g(x) = x2(2x - 3) - x(x + 1) - (3x - 2)
g(x) = 2x3 - 3x2 - x2 - x - 3x + 2
g(x) = 2x3 - 4x2 - 4x + 2
b) f(2) = 2.23 - 4. 22 - 3.2 + 10 = 16 - 16 - 6 + 10 = 4
g(-2) = 2.(-2)3 - 4.(-2)2 - 4.(-2) + 2 = 2 . 8 - 4.4 + 8 + 2 = 10
c) H(x) = f(x) - g(x) = (2x3 - 4x2 - 3x + 10) - (2x3 - 4x2 - 4x + 2)
H(x) = 2x3 - 4x2 - 3x + 10 - 2x3 + 4x2 + 4x - 2
H(x) = (2x3 - 2x3) - (4x2 - 4x2) - (3x - 4x) + (10 - 2)
H(x) = x + 8
=> f(x) - g(x) = A(x) = -x - 8
d) Ta có: H(x) = 0
=> x + 8 = 0
=> x = -8
a, \(f\left(x\right)=2x^2+6x^4-3x^3+2011\)
\(=6x^4-3x^3+2x^2+2011\)
\(g\left(x\right)=2x^3-5x^2-3x^4-2012\)
\(=-3x^4+2x^3-5x^2-2012\)
b, \(f\left(x\right)+g\left(x\right)=6x^4-3x^3+2x^2+2011-3x^4+2x^3-5x^2-2012\)
\(=\left(6x^4-3x^4\right)+\left(2x^3-3x^3\right)+\left(2x^2-5x^2\right)+\left(2011-2012\right)\)
\(=3x^4-x^3-3x^2-1\)
\(f\left(x\right)-g\left(x\right)=6x^4-3x^3+2x^2+2011-\left(-3x^4+2x^3-5x^2-2012\right)\)
\(=6x^4-3x^3+2x^2+2011+3x^4-2x^3+5x^2+2012\)
\(=\left(6x^4+3x^4\right)-\left(3x^3+2x^3\right)+\left(2x^2+5x^2\right)+\left(2011+2012\right)\)
\(=9x^4-5x^3+7x^2+4023\)
f(x) = anxn + an – 1xn– 1 + … + a1x + ao
+
g(x) = bnxn + bn – 1xn– 1 + … + b1x + bo
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f(x) + g(x) = (an + bn)xn + (an – 1 + bn – 1)xn– 1 + ….. + (a1 + b1)x + (ao + bo)