Bài tập 6.11 trang 7 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

a) Mẫu thức chung là: 42x²y⁵

Ta có: 42x²y⁵ : 14x²y = 3y⁴ ; 42x²y⁵ : 21xy⁵ = 2x

Quy đồng mẫu thức ta có:

$\frac{25}{14x^{2}y}=\frac{25.3y^4}{14x^{2}y.3y^4}=\frac{75y^4}{42x^2{y}^5}$

$\frac{14}{21x{y}^5}=\frac{14.2x}{21x{y}^5.2x}=\frac{28x}{42x^2{y}^5}$.

b) Ta có $\frac{4x-4}{2x\left(x+3\right)}=\frac{2\left(2x-2\right)}{2x\left(x+3\right)}=\frac{2x-2}{x\left(x+3\right)}$.

Mẫu thức chung: 3x(x + 3)(x + 1).

Ta có: 3x(x + 3)(x + 1) : x(x + 3) = 3(x + 1)

3x(x + 3)(x + 1) : 3x(x + 1) = (x + 3)

Quy đồng mẫu thức ta có:

$\frac{2x-2}{x\left(x+3\right)}=\frac{\left(2x-2\right).3\left(x+1\right)}{x\left(x+3\right).3\left(x+1\right)}=\frac{3\left(2x-2\right)\left(x+1\right)}{3x\left(x+3\right)\left(x+1\right)}$

$=\frac{6\left(x-1\right)\left(x+1\right)}{3x\left(x+3\right)\left(x+1\right)}=\frac{6\left(x^2-1\right)}{3x\left(x+3\right)\left(x+1\right)}$;

$\frac{x-3}{3x\left(x+1\right)}=\frac{\left(x-3\right).\left(x+3\right)}{3x\left(x+1\right).\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{3x\left(x+3\right)\left(x+1\right)}=\frac{x^2-9}{3x\left(x+3\right)\left(x+1\right)}$.

-- Mod Toán 8 HỌC247