Question

Rút gọn các biểu thức sau:

a) \(\sqrt{50}-\sqrt{18}\) 

b) \(\left(\dfrac{2}{a^2+a}-\dfrac{2}{a+1}\right):\dfrac{1-a}{a^2+2a+1}\)  (với a\(\ne0\) và a\(\ne\pm1\)).

Answer 1

b. \(=\left(\dfrac{2}{a\left(a+1\right)}-\dfrac{2}{a+1}\right):\dfrac{1-a}{a^2+2a+1}\)

\(=\left(\dfrac{2-2a}{a\left(a+1\right)}\right):\dfrac{1-a}{\left(a+1\right)^1}\)

\(=\dfrac{\left(2-2a\right)\left(a+1\right)^2}{a\left(a+1\right)\left(1-a\right)}\)

\(=\dfrac{2\left(1-a\right)\left(a+1\right)^2}{a\left(a+1\right)\left(1-a\right)}=\dfrac{2\left(a+1\right)}{a}\)

Answer 2

a.\(=\sqrt{2}.\left(\sqrt{25}-\sqrt{9}\right)=\sqrt{2}.\left(5-3\right)=2\sqrt{2}\)

 

Answer 3

a: \(=5\sqrt{2}-3\sqrt{2}=2\sqrt{2}\)

b: \(=\dfrac{2-2a}{a\left(a+1\right)}\cdot\dfrac{\left(a+1\right)^2}{1-a}=\dfrac{2a+2}{a}\)

Answer 4

\(b.\left(\dfrac{2}{a^2+a}-\dfrac{2}{a+1}\right):\dfrac{1-a}{a^2+2a+1}\left(đk:a\ne0;a\ne\pm1\right)\)

\(=\left[\dfrac{2\left(a+1\right)}{a\left(a+1\right)^2}-\dfrac{2a\left(a+1\right)}{a\left(a+1\right)^2}\right].\dfrac{a^2+2a+1}{1-a}\)

\(=\dfrac{2a+2-2a^2-2a}{a\left(a+1\right)^2}.\dfrac{\left(a+1\right)^2}{1-a}\)

\(=\dfrac{2\left(1+a\right)\left(1-a\right)}{a\left(a+1\right)^2}.\dfrac{\left(a+1\right)^2}{1-a}\)

\(=\dfrac{2\left(1+a\right)}{a}\)

\(=\dfrac{2+2a}{a}\)