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b) ĐK: \(a\ge0,a\ne6\)
\(\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\frac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}-6\)
=\(\left(3-\sqrt{a}\right)-\left(\sqrt{a}-3\right)-6=3-\sqrt{a}-\sqrt{a}+3-6\)
\(=-2\sqrt{a}\)
\(a,\)\(\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right).\left(2-\frac{\sqrt{a}\left(b+5\right)}{\sqrt{b}-5}\right).\)
\(=\left(2-\sqrt{a}\right)\left(\frac{2\sqrt{b}-10-\sqrt{ab}-5\sqrt{a}}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(\frac{2\left(\sqrt{b}-5\right)-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\frac{\left(2-\sqrt{a}\right)\left(2-\sqrt{a}\right)\left(\sqrt{b}-5\right)}{\sqrt{b}-5}=\left(2-\sqrt{a}\right)^2\)
\(=a-4\sqrt{a}+4\)
\(b,\frac{9-a}{\sqrt{a}+3}-\frac{9-6\sqrt{a}+a}{\sqrt{a}-3}-6\)
\(=\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\frac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}-6\)
\(=3-\sqrt{a}-\left(\sqrt{a}-3\right)-6\)
\(=-2\sqrt{a}\)
\(b,\frac{9-a}{\sqrt{a}+3}-\frac{9-6\sqrt{a}+a}{\sqrt{a}-3}-6\)
\(=\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\frac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}-6\)
\(=3-\sqrt{a}-\left(\sqrt{a}-3\right)-6\)
\(=3-\sqrt{a}-\sqrt{a}+3-6\)
\(=-2\sqrt{a}\)
a) \(\left(2-\frac{a-3.\sqrt{a}}{\sqrt{a}-3}\right).\left(2-\frac{5.\sqrt{a}+\sqrt{a}.b}{\sqrt{b}-5}\right)\)
=\(\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2+\frac{\sqrt{a}\left(5-\sqrt{b}\right)}{5-\sqrt{b}}\right)\)
=\(\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)\)
=4-a ( Bạn xem lại đề bài giúp mình )
b)\(\frac{9-a}{\sqrt{a}+3}-\frac{9-6\sqrt{a}+a}{\sqrt{a}-3}\) -6
=\(\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}+\frac{\left(3-\sqrt{a}\right)^2}{3-\sqrt{a}}-6\)
=\(3-\sqrt{a}+3-\sqrt{a}-6\)
=-2\(\sqrt{a}\)