bài 1: Chứng minh rằng:
a) \(\left(4+\sqrt{15}\right)\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)=2\)
b) \(\dfrac{1}{\sqrt{\left(\sqrt{6}-1\right)^2+1}}-\dfrac{1}{\sqrt{\left(\sqrt{6}+1\right)^2-1}}=0\)
c) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}=x-y\) với x > 0;y > 0;x ≠ y
d) \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}=\dfrac{4\sqrt{a}}{a-1}=\dfrac{2}{\sqrt{x}-4}\) với a ∈ R,a ≥ 0 và a ≠ 1
e) \(\left(\dfrac{1}{\sqrt{x}+4}+\dfrac{1}{\sqrt{x}-4}\right)\dfrac{\sqrt{x}+4}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-4}\) với x ≥ 0;x ≠ 16
giúp với ạ
`a.`\(\left(4+\sqrt{15}\right)\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right).1}\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}+\sqrt{3}\right|.\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}+\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\)
`b.` Đề sai
\(\dfrac{1}{\sqrt{\left(\sqrt{6}-1\right)^2+1}}-\dfrac{1}{\sqrt{\left(\sqrt{6}+1\right)^2-1}}=0\)
\(\Leftrightarrow\left(\sqrt{6}-1\right)^2+1=\left(\sqrt{6}+1\right)^2-1\)
\(\Leftrightarrow6-2\sqrt{6}+1+1=6+2\sqrt{6}+1-1\)
\(\Leftrightarrow4\sqrt{6}=2\) ( vô lý )
`c.`\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
`d.`\(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1}{a-1}\)
\(=\dfrac{4\sqrt{a}}{a-1}\)
`e.`\(\left(\dfrac{1}{\sqrt{x}+4}+\dfrac{1}{\sqrt{x}-4}\right).\dfrac{\sqrt{x}+4}{\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}-4+\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right).\dfrac{\sqrt{x}+4}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}.\dfrac{\sqrt{x}+4}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}-4}\)