Question

Bài 1:

a, \(\frac{x}{4x^2y^3}và\frac{5}{6x^3y}\)

b, \(\frac{2}{x^2+2xy}và\frac{1}{xy+2y^2}\)

c, \(\frac{5}{6-2x}và\frac{3}{x^2-9}\)

Answer 1

\(a.\frac{x}{4x^2y^3}và\frac{5}{6x^3y}\)

- MTC: 12x³y³

- NTP: 3x ; 2y²

^{\(\frac{x}{4x^2y^3}=\frac{x.3x}{4x^2y^3.3x}=\frac{3x^2}{12x^3y^3}\\ \frac{5}{6x^3y}=\frac{5.2y^2}{6x^3y.2y^2}=\frac{10y^2}{12x^3y^3}\)}

^{\(b.\frac{2}{x^2+2xy}và\frac{1}{xy+2y^2}\)}

- Ta có : x² + 2xy = x ( x + 2y ) ; xy + 2y² = y (x + 2y ).

-MTC : xy(x+2y)

- NTP : y ; x.

\(\frac{2}{x^2+2xy}=\frac{2.y}{y\left(x^2+2xy\right)}=\frac{2y}{x^2y+2xy^2}=\frac{2y}{xy\left(x+2y\right)}\\ \frac{1}{xy+2y^2}=\frac{1.x}{x\left(xy+2y^2\right)}=\frac{x}{x^2y+2xy^2}=\frac{x}{xy\left(x+2y\right)}\)

\(c.\frac{5}{6-2x}và\frac{3}{x^2-9};\frac{-5}{2x-6}và\frac{3}{x^2-9}\)

- Ta có : 2x - 6 = 2 ( x - 3 ) ; x² - 9 = ( x + 3 ) ( x - 3 )

- MTC : 2 ( x - 3 ) ( x + 3 )

- NTP : x + 3 ; 2.

\(\frac{-5}{2x-6}=\frac{-5\left(x+3\right)}{\left(2x-6\right)\left(x+3\right)}=\frac{-5x-15}{2\left(x-3\right)\left(x+3\right)}\\ \frac{3}{x^2-9}=\frac{3.2}{2\left(x^2-9\right)}=\frac{6}{2\left(x-3\right)\left(x+3\right)}\)

Answer 2

Câu hỏi là gì thế bạn??????

Answer 3

Quên chưa có đề bài

Answer 4

Đề: Quy đồng mẫu thức hai phân thức