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\(a,A\left(x\right)=2x^3-3x^2+2x+1\\ B\left(x\right)=3x^3+2x^2-x-5\\ b,A\left(x\right)+B\left(x\right)=\left(2x^3+3x^3\right)+\left(2x^2-3x^2\right)+\left(2x-x\right)+\left(1-5\right)=5x^3-x^2+x-4\\ A\left(x\right)-B\left(x\right)=\left(2x^3-3x^3\right)+\left(-3x^2-2x^2\right)+\left(2x+x\right)+\left(1-5\right)=-x^3-5x^2+3x-4\)
a) Ta có : \(A\left(x\right)+B\left(x\right)\)
\(=2x^3+2x-3x^2+1+2x^2+3x^3-x-5\)
\(=\left(2x^3+3x^3\right)+\left(-3x^2+2x^2\right)+\left(2x-x\right)+\left(1-5\right)\)
\(=5x^3-x^2-x-4\)
b) Ta sẽ sắp xếp như sau :
\(A\left(x\right)=2x^3-3x^2+2x+1\)
\(B\left(x\right)=3x^3+2x^2-x-5\)
c) Ta có : \(A\left(x\right)-B\left(x\right)\)
\(=\left(2x^3+2x-3x^2+1\right)-\left(2x^2+3x^3-x-5\right)\)
\(=2x^3+2x-3x^2+1-2x^2-3x^3+x+5\)
\(=\left(2x^3-3x^3\right)+\left(-3x^2-2x^2\right)+\left(2x+x\right)+\left(1+5\right)\)
\(=-x^3-5x^2+3x+6\)
a)
\(A\left(x\right)=-2x+7+2x^2\\ \text{ }=2x^2-2x+7\)
b)
\(A\left(x\right)+B\left(x\right)=\left(2x^2-2x+7\right)+\left(x^2+2x-2\right)\\ \text{ }=2x^2-2x+7+x^2+2x-2\\ \text{ }=\left(2x^2+x^2\right)+\left(2x-2x\right)+\left(7-2\right)\\ \text{ }=3x^2+5\)
c)
\(C\left(x\right)\cdot B\left(x\right)=x\cdot\left(x^2+2x-2\right)\\ \text{ }=x\cdot x^2+x\cdot2x+x\cdot\left(-2\right)\\ \text{ }=x^3+2x^2-2x\)
A(X)=2x2+2x-3x2+1
=-x2+2x+1
B(x)=2x2+3x3-x-5
=3x3+2x2-x-5
A(X)+B(x)=-x2+2x+1+3x3+2x2-x-5=3x3+x2+x-4
A(X)-B(x)=-x2+2x+1-3x3-2x2+x+5=-3x3-3x2+3x+6
a: \(P\left(x\right)=5x^5-4x^4+2x^2+3x+6\)
\(Q\left(x\right)=-x^5+2x^4-2x^3+3x^2+x+\dfrac{1}{4}\)
b: \(P\left(x\right)+Q\left(x\right)=4x^5-2x^4-2x^3+5x^2+4x+\dfrac{25}{4}\)
a: A(x)=2x^3+x^2+4x+1
B(x)=-2x^3+x^2+3x+2
b: M(x)=A(x)+B(x)
=2x^3+x^2+4x+1-2x^3+x^2+3x+2
=2x^2+7x+3
c: M(x)=0
=>2x^2+7x+3=0
=>2x^2+6x+x+3=0
=>(x+3)(2x+1)=0
=>x=-3 hoặc x=-1/2
A(x)2x3 2x3x2 1
B(x)2x2 3x3 x5
Dấu gì?
A(x)=2xmũ3 + 2x - 3xmũ2 +1 B(x)=2x mũ2 + 3xmũ3 - x - 5