Question

Thu gọn đa thức, tìm bậc, hệ số cao nhất của đa thức.

C = x²y - 1/2xy² + 1/3x²y + 2/3xy² + 1

D = xy^2z + 3xyz² - 1/5xy²z - 1/3xyz² - 2

E = 3xy⁵ - x²y + 7xy - 3xy⁵ + 3x²y - 1/2xy + 1

K = 5x³ - 4x + 7x² - 6x³ + 4x + 1

F = 12x³y² - 3/7x⁴y² + 2xy³ - x³y² + x⁴y² - xy³ - 5

Answer 1

C= x² y - \(\dfrac{1}{2}\)xy² + \(\dfrac{1}{3}\)x²y +\(\dfrac{2}{3}\)xy² + 1

C=(x²y + \(\dfrac{1}{3}\)x²y )+( - \(\dfrac{1}{2}\)xy² +\(\dfrac{2}{3}\)xy²)+ 1

C=\(\dfrac{4}{3}\)x²y +\(\dfrac{1}{6}\)xy²+1

=>Bặc: 3

D= xy²z + 3xyz² - \(\dfrac{1}{5}\)xy²z - \(\dfrac{1}{3}\)xyz² - 2

D=(xy²z - \(\dfrac{1}{5}\)xy²z )+( 3xyz² - \(\dfrac{1}{3}\)xyz²) - 2

D=\(\dfrac{4}{5}\)xy²z +\(\dfrac{8}{3}\)xyz² - 2

=> Bậc :4

E = 3xy⁵ - x²y + 7xy - 3xy⁵ + 3x²y - \(\dfrac{1}{2}\)xy + 1

E=(3xy⁵- 3xy⁵) + (- x²y + 3x²y) + (7xy - \(\dfrac{1}{2}\)xy)+ 1

E= 2x²y + \(\dfrac{13}{2}\)xy + 1

=> Bậc: 3

K = 5x³ - 4x + 7x² - 6x³ + 4x + 1

K= (5x³ - 6x³ ) + (- 4x + 4x) +1

K= -1x³ + 1

=>Bậc: 3

F = 12x³y² - \(\dfrac{3}{7}\)x⁴y² + 2xy³ - x³y² + x⁴y² - xy³ - 5

F=( 12x³y² - x³y²) + (- \(\dfrac{3}{7}\)x⁴y² + x⁴y²) + (2xy³ - xy³) -5

F=11x³y² + \(\dfrac{4}{7}\)x⁴y² + xy³ - 5

=> Bậc :6

CHÚC BN HỌC TỐT ^-^