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f(x) + g(x) = 2x4 + 2x2
f(x) - g(x) = x4 - x2 + 2x
suy ra : f(x) = [ ( 2x4 + 2x2 ) + ( x4 - x2 + 2x ) ] : 2 = \(\frac{3x^4+x^2+2x}{2}\)
g(x) = [ ( 2x4 + 2x2 ) - ( x4 - x2 + 2x ) ] : 2 = \(\frac{x^4+3x^2-2x}{2}\)
Ta có:\(f\left(x\right)-h\left(x\right)=g\left(x\right)\Leftrightarrow h\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(\Leftrightarrow h\left(x\right)=\left(2x^4+5x^3-x+8\right)-\left(x^4-x^2+3x+9\right)\)
\(=2x^4+5x^3-x+8-x^4-x^2-3x-9\)
\(=x^4+5x^3+x^2-4x-1.\)
Vậy, đa thức cần tìm là: \(h\left(x\right)=x^4+5x^3+x^2-4x-1.\)
Ta có: \(h\left(x\right)-g\left(x\right)=f\left(x\right)\Leftrightarrow h\left(x\right)=f\left(x\right)+g\left(x\right)\)
\(\Leftrightarrow h\left(x\right)=\left(2x^4+5x^3-x+8\right)+\left(x^4-x^2+3x+9\right)\)
\(=2x^4+5x^3-x+8+x^4-x^2+3x+9\)
\(=3x^4+5x^3-x^2+2x+17\)
Vậy, đa thức cần tìm là:\(h\left(x\right)=3x^4+5x^3-x^2+2x+17.\)
Bài 1:
a) Ta có: \(P\left(x\right)=3x^4+2x^2-3x^4-2x^2+2x-5\)
\(=\left(3x^4-3x^4\right)+\left(2x^2-2x^2\right)+2x-5\)
\(=2x-5\)
Bài 1:
b)
\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)
\(P\left(3\right)=2\cdot3-5=6-5=1\)
Xét [\(f\left(x\right)+g\left(x\right)\)]+[\(f\left(x\right)-g\left(x\right)\)]=\(\left[2x^4+5x^2-3x\right]\)+\(\left[x^4-x^2+2x\right]\)
\(2f\left(x\right)=2x^4+5x^2-3x+x^4-x^2+2x\)
\(2f\left(x\right)=3x^4+4x^2-x\)
\(\Rightarrow f\left(x\right)=\dfrac{3x^4+4x^2-x}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^4+2x^2-\dfrac{1}{2}x\)
Xét \(\left[f\left(x\right)+g\left(x\right)\right]-\left[f\left(x\right)-g\left(x\right)\right]=\)\(\left[2x^4+5x^2-3x\right]\)\(-\)\(\left[x^4-x^2+2x\right]\)
\(2g\left(x\right)=\)\(2x^4+5x^2-3x-x^4+x^2-2x\)
\(2g\left(x\right)=x^4+6x^2-5x\)
\(\Rightarrow g\left(x\right)=\dfrac{x^4+6x^2-5x}{2}\)
\(\Rightarrow g\left(x\right)=\dfrac{1}{2}x^4+3x^2-\dfrac{5}{2}x\)