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Chhi Phuong

10/08/2017 at 21:52
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Cho đa thức: f(x)= 10x5−8x4+6x3−4x2+2x+210x5−8x4+6x3−4x2+2x+2

g(x)=−5x5+4x4−3x3+3x2−5x+2−5x5+4x4−3x3+3x2−5x+2

h(x)=−x5+2x4−x3+x−7−x5+2x4−x3+x−7

a) Tính f(x)+g(x)-h(x) và f(x)-g(x)-h(x).Tìm bậc,hệ số cao nhất và hệ số tự do của đa thức kết quả.

b)Tìm x để f(x)+2g(x)=0




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    Phan Thanh Tinh Coordinator 11/08/2017 at 00:10

    a) \(f\left(x\right)=10x^5-8x^4+6x^3-4x^2+2x+2\)

    \(g\left(x\right)=-5x^5+4x^4-3x^3+3x^2-5x+2\)

    \(h\left(x\right)=-x^5+2x^4-x^3+x-7\)

    \(f\left(x\right)+g\left(x\right)-h\left(x\right)=6x^5-6x^4+4x^3-x^2-4x+11\)

    \(f\left(x\right)-g\left(x\right)-h\left(x\right)=16x^5-14x^4+10x^3-7x^2+6x+7\)

    The degrees of 2 polynomials are both 5

    b) \(f\left(x\right)+2g\left(x\right)=0\)

    \(\Leftrightarrow10x^5-8x^4+6x^3-4x^2+2x+2+2\left(-5x^5+4x^4-3x^3+3x^2-5x+2\right)=0\)

    \(\Leftrightarrow2x^2-8x+6=0\Leftrightarrow x^2-4x+3=0\)

    \(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

    P/S : I forgot the way to translate "hệ số tự do" and "hệ số cao nhất" into English. I'll come back soon 


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