Bài 6 :
a) \(...\Leftrightarrow A=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\dfrac{4}{\sqrt{x}+2}\)
b) \(...\Leftrightarrow B=\dfrac{\left(x-\sqrt{2024}\right)\left(x+\sqrt{2024}\right)}{\left(x+\sqrt{2024}\right)^2}\) Sửa lại tử số là \(x^2-2024\)
\(\Leftrightarrow B=\dfrac{x-\sqrt{2024}}{x+\sqrt{2024}}=\dfrac{x+\sqrt{2024}-2\sqrt{2024}}{x+\sqrt{2024}}=1-\dfrac{2}{x+\sqrt{2024}}\)
Bài 7 :
a) \(...\Leftrightarrow A=5x-5\sqrt{5}+\dfrac{\sqrt{x^2\left(x+5\right)}}{\sqrt{x+5}}\)
\(\Leftrightarrow A=5x-5\sqrt{5}+\dfrac{\left|x\right|\sqrt{x+5}}{\sqrt{x+5}}\)
\(\Leftrightarrow A=5x-5\sqrt{5}+x\left(x\ge0\right)\)
\(\Leftrightarrow A=6x-5\sqrt{5}\)
\(\Leftrightarrow A=6\sqrt{5}-5\sqrt{5}=\sqrt{5}\left(với.x=\sqrt{5}\right)\)
b) \(...\Leftrightarrow B=\sqrt{\left(a^2-1\right)+2\sqrt{\left(a^2-1\right)}+1}-\sqrt{\left(a^2-1\right)-2\sqrt{\left(a^2-1\right)}+1}\)
\(\Leftrightarrow B=\sqrt{\left(\sqrt{a^2-1}+1\right)^2}-\sqrt{\left(\sqrt{a^2-1}-1\right)^2}\)
\(\Leftrightarrow B=\left|\sqrt{a^2-1}+1\right|-\left|\sqrt{a^2-1}-1\right|\)
\(\Leftrightarrow B=\left|\sqrt{\left(\sqrt{5}\right)^2-1}+1\right|-\left|\sqrt{\left(\sqrt{5}\right)^2-1}-1\right|\left(với.a=\sqrt{5}\right)\)
\(\Leftrightarrow B=\left|2+1\right|-\left|2-1\right|=2\)
