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`a,f(x)-g(x)+h(x)`
`=x^3-2x^2+3x+1-(x^3+x-1)+2x^2-1`
`=(x^3-x^3)+(2x^2-2x^2)+3x+1+1-1`
`=0+0+3x+1`
`=3x+1`
`b,f(x)-g(x)+h(x)=0`
`=>3x+1=0`
`=>x=-1/3`
\(\text{a)}f\left(x\right)-g\left(x\right)+h\left(x\right)=\left(x^3-2x^2+3x+1\right)-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
\(=x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)
\(=\left(x^3-x^3\right)+\left(-2x^2+2x^2\right)+\left(3x-x\right)+\left(1+1-1\right)\)
\(=2x+1\)
\(\text{b)Vì f(x)-g(x)+h(x)=0}\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x\) \(=0-1=-1\)
\(\Rightarrow\) \(x\) \(=\left(-1\right):2=\dfrac{-1}{2}\)
\(\text{Vậy x=}\dfrac{-1}{2}\text{ thì f(x)-g(x)+h(x)=0}\)
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: \(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)
b: \(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)
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: \(F\left(x\right)=x^3+2x^2+3x+4\)
\(G\left(x\right)=x^3-x^2+3x+1\)
b: \(F\left(x\right)+G\left(x\right)=2x^3+x^2+6x+5\)
\(F\left(x\right)-G\left(x\right)=3x^2+3\)
a) f(x)-g(x)+h(x)=\(\left(x^3-2x^2+3x+1\right)-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
=\(x^3-2x^2+3x+1-x^3+x-1+2x^2-1\)
=\(4x-1\)
Vậy f(x)-g(x)+h(x)=4x-1
Ta có:f(x)-g(x)+h(x)=4x-1=0
=> 4x-1=0
=> 4x=1
=> x=1/4
Vậy để f(x)-g(x)+h(x)=0 thì x=1/4
b. h(x) = (2x3 + 3x2 - 2x + 3) - (2x3 + 3x2 - 7x + 2)
= 2x3 + 3x2 - 2x + 3 - 2x3 - 3x2 + 7x - 2
= 5x + 1 (0.5 điểm)
g(x) = (2x3 + 3x2 - 2x + 3) + (2x3 + 3x2 - 7x + 2)
= 2x3 + 3x2 - 2x + 3 + 2x3 + 3x2 - 7x + 2
= 4x3 + 6x2 - 9x + 5 (0.5 điểm)
câu 4: b, đề bài là tính giá trị của A tại x =-1/2;y=-1
Tk
Bài 2
a) F(x)-G(x)+H(x)= \(x^3-2x^2+3x+1-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
= \(x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)
= \(x^3-x^3-2x^2+2x^2+3x-x+1+1-1\)
= 2x + 1
b) 2x + 1 = 0
2x = -1
x=\(\dfrac{-1}{2}\)