a. ĐKXĐ: \(x\le1\)
b.
\(f\left(x\right)=\sqrt{1-x+2\sqrt{1-x}+1}=\sqrt{\left(\sqrt{1-x}+1\right)^2}=\sqrt{1-x}+1\)
Xét với \(x_1< x_2\le1\) ta có:
\(f\left(x_1\right)-f\left(x_2\right)=\sqrt{1-x_1}+1-\left(\sqrt{1-x_2}+1\right)=\sqrt{1-x_1}-\sqrt{1-x_2}\)
\(=\dfrac{1-x_1+\left(1-x_2\right)}{\sqrt{1-x_1}+\sqrt{1-x_2}}=\dfrac{x_2-x_1}{\sqrt{1-x_1}+\sqrt{1-x_2}}>0\) (do \(x_1< x_2\))
\(\Rightarrow f\left(x_1\right)>f\left(x_2\right)\Rightarrow\) hàm nghịch biến trên TXĐ










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