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Bài 1:
$14+\sqrt{40}+\sqrt{56}+\sqrt{140}=14+\sqrt{56}+(\sqrt{40}+\sqrt{140})$
=14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}=(12+2\sqrt{35})+2+(2\sqrt{10}+2\sqrt{14})$
$=(\sqrt{5}+\sqrt{7})^2+2+2\sqrt{2}(\sqrt{5}+\sqrt{7})$
$=(\sqrt{5}+\sqrt{7}+\sqrt{2})^2$
$\Rightarrow \sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}=\sqrt{2}+\sqrt{5}+\sqrt{7}$
\(\Rightarrow A=\frac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=1\)
Lời giải:
a) ĐKXĐ: $a,b\geq 0$ và $a,b$ không đồng thời cùng bằng $0$
\(B=\frac{2a+2\sqrt{2}a-2\sqrt{3ab}+2\sqrt{3ab}-3b-2a\sqrt{2}}{a\sqrt{2}+\sqrt{3ab}}=\frac{2a-3b}{\sqrt{a}(\sqrt{2a}+\sqrt{3b})}=\frac{(\sqrt{2a}-\sqrt{3b})(\sqrt{2a}+\sqrt{3b})}{\sqrt{a}(\sqrt{2a}+\sqrt{3b})}\)
\(=\frac{\sqrt{2a}-\sqrt{3b}}{\sqrt{a}}=\sqrt{2}-\sqrt{\frac{3b}{a}}\)
b)
\(a=1+3\sqrt{2}; 3b=30+11\sqrt{8}\Rightarrow \frac{3b}{a}=\frac{30+11\sqrt{8}}{1+3\sqrt{2}}=\frac{(30+11\sqrt{8})(1-3\sqrt{2})}{(1+3\sqrt{2})(1-3\sqrt{2})}\)
\(=\frac{102+68\sqrt{2}}{17}=6+4\sqrt{2}=(2+\sqrt{2})^2\)
\(\Rightarrow \sqrt{\frac{3b}{a}}=2+\sqrt{2}\)
\(\Rightarrow B=\sqrt{2}-(2+\sqrt{2})=-2\)
a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
\(S=\frac{\left[\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(S=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^2\sqrt{a}+\left(\sqrt{b}\right)^2\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{b}-\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(S=\frac{\left(\sqrt{a}\right)^3-3\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{3\left(\sqrt{a}\right)^3-3a\sqrt{b}+3\sqrt{a}b}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\sqrt{a}\left[\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3\right]}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)