Question

A=\(\dfrac{\sqrt{x}}{\sqrt{x}+4}\) + \(\dfrac{4}{\sqrt{x}-4}\) (x>=0;x khac 10)

C=\(\dfrac{2}{x-1}\) + \(\dfrac{1}{\sqrt{x}+1}\) (x>=0;x khac 1)

giup mik vs

 

Answer 1

Đề yêu cầu gì em nhỉ?

Answer 2

rut gon thoi a

Answer 3

\(A=\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\dfrac{4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\)

\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{x+16}{x-16}\)

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

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

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

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