Question

Cho các đa thức C= x\(^5\) - 2x\(^3\) + 3x\(^4\) - 9x\(^2\) +11x - 6

và B=3x\(^4\) + x\(^5\) - 2(x\(^3\) + 4)- 10x\(^2\) + 9x

Tính M=B-C. Tìm x để M=2x + 2.

Answer 1

sắp xếp:
C= \(x^5\) + 3\(x^4\) - 2\(x^3\) - 9\(x^2\) + 11x - 6

B= \(x^5\) + \(3x^4\) - \(2x^3\) - \(10x^2\) +9x + 4

B= \(x^5\) + \(3x^4\) - \(2x^3\) - \(10x^2\) +9x + 4
+
- C= \(x^5\) - 3\(x^4\) + 2\(x^3\) + 9\(x^2\) - 11x + 6

M = \(2x^5\) - \(x^2\) - 2x + 10

Answer 2

Ta có M = B - C

\(\Rightarrow M=[3x^4+x^5-2\left(x^3+4\right)-10x^2+9x]\\ \\ -\left(x^5-2x^3+3x^4-9x^2+11x-6\right)\)

\(\Rightarrow M=3x^4+x^5-2x^3+4-10x^2+9x\\ -x^5+2x^3-3x^4+9x^2-11x+6\)

\(\Rightarrow M=\left(3x^4-3x^4\right)+\left(x^5-x^5\right)+\left(-2x^3+2x^3\right)\\ +\left(4+6\right)+\left(-10x^2+9x^2\right)+\left(9x-11x\right)\)

\(\Rightarrow M=10-x^2-2x\)

Vậy \(M=10-x^2-2x\)

Answer 3

Hồi nãy mình tính nhầm

Ta có M = B - C

\(\Rightarrow M=[3x^4+x^5-2\left(x^3+4\right)-10x^2+9x]-\\ \left(x^5-2x^3+3x^4-9x^2+11x-6\right)\)

\(\Rightarrow M=3x^4+x^5-2x^3+8-10x^2+9x\\ -x^5+2x^3-3x^4+9x^2-11x+6\)

\(\Rightarrow M=\left(3x^4-3x^4\right)+\left(x^5-x^5\right)+\left(-2x^3+2x^3\right)\\ +\left(8+6\right)+\left(9x-11x\right)+\left(-10^2+9x^2\right)\)

\(\Rightarrow M=14-2x-x^2\)