\(ĐKXĐ:a\ge0\)

\(A=\left(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\right):\left(1+\frac{\sqrt{a}}{a+1}\right)\)

\(\Leftrightarrow A=\left(\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}+\frac{1}{\sqrt{a}+1}\right):\frac{a+\sqrt{a}+1}{a+1}\)

\(\Leftrightarrow A=\frac{2\sqrt{a}+a+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a+1}{a+\sqrt{a}+1}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}\)

\(\Leftrightarrow A=\frac{\sqrt{a}+1}{a+\sqrt{a}+1}\)

Bài 1:

\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)

\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)

\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)

\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)

\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)

\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)

\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)

\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)

\(\Rightarrow C=\sqrt{14}\)

\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)

\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)

\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)

\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)

\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)

Bài 2:

a) Bạn xem lại đề.

b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)

c)

\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)

\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)

a: \(A=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)

\(=2+\dfrac{2a+2}{\sqrt{a}}=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

b: Để A=7 thì \(2a-5\sqrt{a}+2=0\)

\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)

=>a=4 hoặc a=1/4

a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)

c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)

Rút gọn

a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)

\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)

\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)

\(=\sqrt{2}\left(1+1\right)=2\sqrt{2}\)

b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)

\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)

\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)

\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)

\(=4\sqrt{a}-5\sqrt{10a}\)

c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)

\(=6+\sqrt{15}-\sqrt{60}\)

\(=6-\sqrt{15}\)