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\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)
\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)
\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)
Điều kiện : a, b\(\ge0\)
\(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\div\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)
\(=\left(\frac{\sqrt{a}.\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)
\(=\left(\frac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a.\sqrt{a}.\left(\sqrt{b}-\sqrt{a}\right)+b.\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right).\left(b-a\right)}{\sqrt{ab}.\left(b-a\right)}\right)\)
\(=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a\sqrt{ab}-a^2+b\sqrt{ab}+b^2-b^2+a^2}{\sqrt{ab}.\left(b-a\right)}\right)\)
giải tiếp
\(=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a\sqrt{ab}+b\sqrt{ab}}{\sqrt{ab}\left(b-a\right)}\right)\)
\(=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{\sqrt{ab}.\left(a+b\right)}{\sqrt{ab}.\left(b-a\right)}\right)=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right).\left(\frac{b-a}{a+b}\right)\)
\(=\frac{b-a}{\sqrt{a}+\sqrt{b}}=\frac{\left(b-a\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b}=\frac{b\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}}{a-b}\)
A=(2+\(\frac{3+\sqrt{3}}{\sqrt{3}+1}\)) . (2-\(\frac{3-\sqrt{3}}{\sqrt{3}-3}\))
=(\(2+\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\)) . (\(2-\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\))
=(\(2+\sqrt{3}\)) . (\(2-\sqrt{3}\))
=22-(\(\sqrt{3}\))2=4-3=1
B=(\(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\)) . (\(a\sqrt{b}-b\sqrt{a}\))
=(\(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\)) . (\(a\sqrt{b}-b\sqrt{a}\))
=(\(\frac{\sqrt{b}.\sqrt{b}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}.\sqrt{a}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\)). (a\(\sqrt{b}-b\sqrt{a}\))
=\(\frac{b-a}{\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)}.\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
=b-a
Ta có: \(A=\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\cdot\left(2-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
\(=\frac{2\left(\sqrt{3}+1\right)+3+\sqrt{3}}{\sqrt{3}+1}\cdot\frac{2\left(\sqrt{3}-1\right)-3+\sqrt{3}}{\sqrt{3}-1}\)
\(=\frac{2\sqrt{3}+2+3+\sqrt{3}}{\sqrt{3}+1}\cdot\frac{2\sqrt{3}-2-3+\sqrt{3}}{\sqrt{3}-1}\)
\(=\frac{3\sqrt{3}+5}{\sqrt{3}+1}\cdot\frac{3\sqrt{3}-5}{\sqrt{3}-1}\)
\(=\frac{2}{2}=1\)
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm