Câu hỏi của Linh nguyễn

\(\sqrt{x}+2\sqrt{x+3}+\sqrt{x^2+3}-7=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)+2\left(\sqrt{x+3}-2\right)+\left(\sqrt{x^2+3}-2\right)=0\)\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2\left(x-1\right)}{\sqrt{x+3}+2}+\dfrac{\left(x+1\right)\left(x-1\right)}{\sqrt{x^2+3}-2}=0\)\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x+3}+2}+\dfrac{x+1}{\sqrt{x^2+3}-2}\right)=0\)\(\Leftrightarrow x=1\)Câu trả lời: \(\sqrt{x}+2\sqrt{x+3}+\sqrt{x^2+3}-7=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)+2\left(\sqrt{x+3}-2\right)+\left(\sqrt{x^2+3}-2\right)=0\)\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2\left(x-1\right)}{\sqrt{x+3}+2}+\dfrac{\left(x+1\right)\left(x-1\right)}{\sqrt{x^2+3}-2}=0\)\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x+3}+2}+\dfrac{x+1}{\sqrt{x^2+3}-2}\right)=0\)\(\Leftrightarrow x=1\)