Question

Cho hai đơn thức sau :

P(x)=5x^5+3x-4x^4-2x^3+6+4x²

Q(x)=2x^4-x+3x²-2x^3+1/4-x^5

Tính:1/3 P(x)-3/4 Q(x)

Answer 1

\(\dfrac{1}{3}P\left(x\right)=\dfrac{5}{3}x^5-\dfrac{4}{3}x^4-\dfrac{2}{3}x^3+\dfrac{4}{3}x^2+x+2\)

\(\dfrac{3}{4}Q\left(x\right)=-\dfrac{3}{4}x^5+\dfrac{3}{2}x^4-\dfrac{3}{2}x^3+\dfrac{9}{4}x^2-\dfrac{3}{4}x+\dfrac{3}{16}\)

Do đó: \(\dfrac{1}{3}P\left(x\right)-\dfrac{3}{4}Q\left(x\right)=\dfrac{29}{12}x^5-\dfrac{17}{6}x^4+\dfrac{5}{6}x^3-\dfrac{11}{12}x^2+\dfrac{7}{4}x+\dfrac{29}{16}\)