Bài 1:
a) A= 3x\(\left(1-\dfrac{4}{5}x+3x^2\right)\)
b) B= (x+4)\(\left(-2x^2-x+3\right)\)
c) C= (x-1)\(\left(x^2+x\right)\)
d) D= \(\left(12x^4-6x^3-4x^2\right)\) : \(\left(-2x^2\right)\)
e) E= \(\left(3x^5+2x^7-4x^4\right)\) : \(6x^3\)
f) F= \(\left(\dfrac{2}{5}x^5-\dfrac{1}{5}x^3+\dfrac{7}{5}x\right)\) : \(\left(-\dfrac{1}{5}x\right)\)
mik cần gấp
\(A = 3x\left(1 - \frac{4}{5}x + 3x^2\right) = 3x \cdot 1 - 3x \cdot \frac{4}{5}x + 3x \cdot 3x^2 = 3x - \frac{12}{5}x^2 + 9x^3\)
\(B = (x + 4)(-2x^2 - x + 3)\\ = x(-2x^2 - x + 3) + 4(-2x^2 - x + 3)\\ = -2x^3 - x^2 + 3x - 8x^2 - 4x + 12\\ = -2x^3 - 9x^2 - x + 12\)
\(C = (x - 1)(x^2 + x) = x(x^2 + x) - 1(x^2 + x)\\ = x^3 + x^2 - x^2 - x = x^3 - x\)
\(D = {(12x^4 - 6x^3 - 4x^2)}:{(-2x^2)} = -6x^2 + 3x + 2\)
\(E = {(3x^5 + 2x^7 - 4x^4)}:{(6x^3)} = \frac{1}{2}x^2 + \frac{1}{3}x^4 - \frac{2}{3}x\)
\(F = \left(\frac{2}{5}x^5 - \frac{1}{5}x^3 + \frac{7}{5}x\right) : \left(-\frac{1}{5}x\right) = -2x^4 + x^2 - 7\)
