Question

[Ngọc Bảo]

giúp mình bài 15,16 ạ . mình cảm ơn
 

Answer 1

[ILoveMath]

Bài 15:

a) \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\left(2-\dfrac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)

\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)

\(=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}-4\right)}{\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+1+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+4}{\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+2}{x-1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+2}{x-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-1}{x-1}\)

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{1}{\sqrt{x}+1}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 16:

a: Ta có: \(Q=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{14}{9-x}\right)\cdot\dfrac{\sqrt{x}-3}{2}\)

\(=\dfrac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2}\)

\(=\dfrac{2x+22}{\sqrt{x}-3}\cdot\dfrac{1}{2}\)

\(=\dfrac{x+11}{\sqrt{x}-3}\)