5x(4x2−2x+1)−2x(10x2−5x+2)=−36

5x(4x2−2x+1)=20x3−10x2+5x

\(2 x \left(\right. 10 x^{2} - 5 x + 2 \left.\right) = 20 x^{3} - 10 x^{2} + 4 x\)

(20x3−10x2+5x)−(20x3−10x2+4x)=−36

20x3−10x2+5x−20x3+10x2−4x=−36,

x=−36

Vậy x=-36

a. Cộng theo từng hạng tử cùng bậc:

• \(x^{4} + \left(\right. - x^{4} \left.\right) = 0\)
• \(- 5 x^{3}\)
• \(3 x^{2}\)
• \(4 x + 2 x = 6 x\)
• \(- 5 + 1 = - 4\)

Vậy:

\(P \left(\right. x \left.\right) + Q \left(\right. x \left.\right) = - 5 x^{3} + 3 x^{2} + 6 x - 4\)

b. P(x)=R(x)+Q(x)

Suy ra:

\(R \left(\right. x \left.\right) = P \left(\right. x \left.\right) - Q \left(\right. x \left.\right)\)

Ta tính:

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

Đổi dấu đa thức \(Q \left(\right. x \left.\right)\):

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

Gộp lại:

• \(x^{4} + x^{4} = 2 x^{4}\)
• \(- 5 x^{3}\)
• \(- 3 x^{2}\)
• \(4 x - 2 x = 2 x\)
• \(- 5 - 1 = - 6\)

Vậy:

\(R \left(\right. x \left.\right) = 2 x^{4} - 5 x^{3} - 3 x^{2} + 2 x - 6\)

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