nhanh nhanh nhanh nhahhhhhh

Bài 5: 1) Ta có: \[ M - N = (-3x^3y + \frac{5}{7}x^3y) - (-\frac{2}{7}x^3y + 2x^3y) \] \[ = -3x^3y + \frac{5}{7}x^3y + \frac{2}{7}x^3y - 2x^3y \] \[ = (-3 - 2)x^3y + (\frac{5}{7} + \frac{2}{7})x^3y \] \[ = -5x^3y + x^3y \] \[ = -4x^3y \] 2) Ta có: \[ M = -N + A \] \[ -3x^3y + \frac{5}{7}x^3y = -(-\frac{2}{7}x^3y + 2x^3y) + A \] \[ -3x^3y + \frac{5}{7}x^3y = \frac{2}{7}x^3y - 2x^3y + A \] \[ -3x^3y + \frac{5}{7}x^3y - \frac{2}{7}x^3y + 2x^3y = A \] \[ (-3 + 2)x^3y + (\frac{5}{7} - \frac{2}{7})x^3y = A \] \[ -x^3y + \frac{3}{7}x^3y = A \] \[ -\frac{7}{7}x^3y + \frac{3}{7}x^3y = A \] \[ -\frac{4}{7}x^3y = A \] Vậy đơn thức \( A \) là \( -\frac{4}{7}x^3y \). Bài 6: 1) Ta có: \[ C - D = \left(5x^3y - \frac{1}{4}x^3y\right) - \left(x^3y + \frac{3}{8}x^3y\right) \] \[ = 5x^3y - \frac{1}{4}x^3y - x^3y - \frac{3}{8}x^3y \] \[ = \left(5 - \frac{1}{4} - 1 - \frac{3}{8}\right)x^3y \] \[ = \left(5 - 1 - \frac{1}{4} - \frac{3}{8}\right)x^3y \] \[ = \left(4 - \frac{2}{8} - \frac{3}{8}\right)x^3y \] \[ = \left(4 - \frac{5}{8}\right)x^3y \] \[ = \left(\frac{32}{8} - \frac{5}{8}\right)x^3y \] \[ = \frac{27}{8}x^3y \] 2) Ta có: \[ D = -C + B \] \[ x^3y + \frac{3}{8}x^3y = -\left(5x^3y - \frac{1}{4}x^3y\right) + B \] \[ x^3y + \frac{3}{8}x^3y = -5x^3y + \frac{1}{4}x^3y + B \] \[ B = x^3y + \frac{3}{8}x^3y + 5x^3y - \frac{1}{4}x^3y \] \[ B = \left(1 + 5 + \frac{3}{8} - \frac{1}{4}\right)x^3y \] \[ B = \left(6 + \frac{3}{8} - \frac{2}{8}\right)x^3y \] \[ B = \left(6 + \frac{1}{8}\right)x^3y \] \[ B = \frac{49}{8}x^3y \] Bài 7: 1) Ta có: $(x+y)+(x-y)=x+y+x-y=x+x+y-y=2x+0=2x.$ 2) Ta có: $(x+y)-(x-y)=x+y-(x-y)=x+y-x+y=x-x+y+y=0+2y=2y.$