Question

giải phương trình :

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

Answer 1

ĐK: x\(\ge0\)

\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+2}}+\dfrac{1}{\sqrt{x+2}+\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}+\sqrt{x}}=1\Leftrightarrow\dfrac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}+\dfrac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}+\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}=1\Leftrightarrow\sqrt{x+3}-\sqrt{x+2}+\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+1}-\sqrt{x}=\sqrt{x+3}-\sqrt{x}=1\Leftrightarrow\sqrt{x+3}=\sqrt{x}+1\Leftrightarrow x+3=x+2\sqrt{x}+1\Leftrightarrow2=2\sqrt{x}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tm\right)\)Vậy S={1}