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a) \(A+B=2x^3+x^2-4x+x^3+3+6x+3x^3-2x+x^2-5\)
\(=6x^3+2x^2-2\)
b) \(A-B=\left(2x^3+x^2-4x+x^3+3\right)-\left(6x+3x^3-2x+x^2-5\right)\)
\(=-8x+8\)
c) Đặt \(f\left(x\right)=-8x+8\)
Ta có: \(f\left(x\right)=0\Leftrightarrow-8x+8=0\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)là nghiệm của đa thức f(x).
a)\(f\left(x\right)=x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-\frac{1}{4}x+2x-3\)
\(=x^5-x^5+7x^4-9x^3-3x^2+2x^2+x^2-\frac{1}{4}x+2x-3\)
\(=7x^4-9x^3+\frac{7}{4}x-3\)
\(g\left(x\right)=5x^4-x^5+\frac{1}{2}x^2+x^5+x^2-4x^4-2x^3+3x^2+x^3-\frac{1}{4}\)
\(=-x^5+x^5+5x^4-4x^4-2x^3+x^3+\frac{1}{2}x^2+x^2+3x^2-\frac{1}{4}\)
\(=x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}\)
b)\(f\left(1\right)=7.1^4-9.1^3+\frac{7}{4}.1-3=7-9+\frac{7}{4}-3=-\frac{13}{4}\)
\(f\left(-1\right)=7.\left(-1\right)^4-9.\left(-1\right)^3+\frac{7}{4}.\left(-1\right)-3=7+9-\frac{7}{4}-3=\frac{45}{4}\)
\(g\left(1\right)=1^4-1^3+\frac{9}{2}.1^2-\frac{1}{4}=1-1+\frac{9}{2}-\frac{1}{4}=\frac{17}{4}\)
\(g\left(-1\right)=\left(-1\right)^4-\left(-1\right)^3+\frac{9}{2}.\left(-1\right)^2-\frac{1}{4}=1+1+\frac{9}{2}-\frac{1}{4}=\frac{25}{4}\)
c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)
\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)
f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)
\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)
\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)
\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)
\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)
\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)
-Bạn kiểm tra lại đề.
a,
*\(P\left(x\right)\) = \(-3x^2+4x-x^3+x^2+3x-1\)
\(P(x)=-3x^2+7x-x^3-1\)
\(P(x)=-x^3-3x^2+7x-1\)
* \(Q(x)=3x^4-x^2+x^3-2x-1-2x^3\)
\(Q(x)=3x^4-x^2-x^3-2x-1\)
\(Q(x)=3x^4-x^3-x^2-1\)
b, \(M(x)=P(x)-Q(x)\)
\(M(x)=-x^3-3x^2+7x-1-3x^4+x^3+x^2+1\)
\(M(x)=-2x^2+7x-3x^4\)
\(A=x^4+6x^2+8x^3-2x-3\)
\(B=3x^2+x^4+4x^3-3x+5\)\(\Rightarrow2B=6x^2+2x^4+8x^3-6x+10\)
\(\Rightarrow A-2B=x^4+6x^2+8x^3-2x-3\)\(-6x^2-2x^4-8x^3+6x-10\)
\(=-x^4+4x-13\)
Ta có
\(A=x^4+8x^3+6x^2-2x-3\)
\(B=x^4+4x^3+3x^2-3x+5\Rightarrow2B=2x^4+8x^3+6x^2-6x+10\)
\(A-2B=x^4+8x^3+6x^2-2x-3\)\(-2x^4-8x^3-6x^2+6x-10\)
\(A-2B=-x^4+4x-13\)