• KQ: \(2 x^{2} - 3 x + 1\)
• Dư: \(0\)

a)V=x.(x+1).(x-1)

b)V= abh=(4+1).4.(4-1)=60

a)V=x.(x+1).(x-1)

b)V= abh=(4+1).4.(4-1)=60

\(5 x \left(\right. 4 x^{2} - 2 x + 1 \left.\right) = 20 x^{3} - 10 x^{2} + 5 x\) \(2 x \left(\right. 10 x^{2} - 5 x + 2 \left.\right) = 20 x^{3} - 10 x^{2} + 4 x\)

\(\left(\right. 20 x^{3} - 10 x^{2} + 5 x \left.\right) - \left(\right. 20 x^{3} - 10 x^{2} + 4 x \left.\right) = - 36\)

\(20 x^{3} - 10 x^{2} + 5 x - 20 x^{3} + 10 x^{2} - 4 x = - 36\) \(x = - 36\)

\(\)a) Tính \(P \left(\right. x \left.\right) + Q \left(\right. x \left.\right)\)

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

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

b) Tìm \(R \left(\right. x \left.\right)\) sao cho \(P \left(\right. x \left.\right) = R \left(\right. x \left.\right) + Q \left(\right. x \left.\right)\)\(\)

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

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

=0 :)

Giỏi:14 HS;Khá:19 HS;Trung Bình:5 HS;Yếu:2 HS

a)-14:3

b)0

c)9:5