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Ta có:
( a + 1 2 ) 2 = 1 (\sqrt{a}+\frac{1}{2})^{2}=1 ( a + 2 1 ) 2 = 1
Nếu
a + 1 2 = − 1 ⇒ a = − 3 2 \sqrt{a}+\frac{1}{2}=-1\Rightarrow \sqrt{a}=-\frac{3}{2} a + 2 1 = − 1 ⇒ a = − 2 3 (loại)
⇒ a + 1 2 = 1 ⇒ a = 1 2 \Rightarrow \sqrt{a}+\frac{1}{2}=1\Rightarrow \sqrt{a}=\frac{1}{2} ⇒ a + 2 1 = 1 ⇒ a = 2 1 (1)
Lại có:
( b − 1 2 ) 2 = 1 (\sqrt{b}-\frac{1}{2})^{2}=1 ( b − 2 1 ) 2 = 1
Nếu
b − 1 2 = − 1 ⇒ b = − 1 2 \sqrt{b}-\frac{1}{2}=-1\Rightarrow \sqrt{b}=-\frac{1}{2} b − 2 1 = − 1 ⇒ b = − 2 1 (loại)
⇒ b − 1 2 = 1 ⇒ b = 3 2 \Rightarrow \sqrt{b}-\frac{1}{2}=1\Rightarrow \sqrt{b}=\frac{3}{2} ⇒ b − 2 1 = 1 ⇒ b = 2 3 (2)
Từ (1) và (2)
⇒ b − a = 3 2 − 1 2 = 1 \Rightarrow \sqrt{b}-\sqrt{a}=\frac{3}{2}-\frac{1}{2}=1 ⇒ b − a = 2 3 − 2 1 = 1