Lời giải:
\(P=\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}=\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

\(=\sqrt{(7+2\sqrt{7.5}+5)+2(\sqrt{10}+\sqrt{14})+2}\)

\(=\sqrt{(\sqrt{7}+\sqrt{5})^2+2\sqrt{2}(\sqrt{5}+\sqrt{7})+(\sqrt{2})^2}\)

\(=\sqrt{(\sqrt{5}+\sqrt{7}+\sqrt{2})^2}=\sqrt{5}+\sqrt{7}+\sqrt{2}\)

\(P=\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}=\left|\sqrt{2}+\sqrt{5}+\sqrt{7}\right|=\sqrt{2}+\sqrt{5}+\sqrt{7}\)

Ta có

\(P=\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

\(\Leftrightarrow P=\sqrt{\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)^2}\)

\(\Leftrightarrow P=\sqrt{5}+\sqrt{2}+\sqrt{7}\)

Mà \(P=\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{5}+\sqrt{2}+\sqrt{7}\)

Suy ra \(a+b+c=5+2+7=14\)

\(A=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

SBT nha

B=\(\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

B=\(\sqrt{7+5+2+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{7.5}}\)

B=\(\left(\sqrt{\sqrt{7}+\sqrt{5}+\sqrt{2}}\right)^2\)

B=\(\sqrt{7}+\sqrt{5}+\sqrt{2}\)

P=\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)=\(\sqrt{2+5+7+2\sqrt{5.2}+2\sqrt{2.7}+2\sqrt{3.5}}\)

=\(\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}\)=\(\sqrt{2}+\sqrt{5}+\sqrt{7}\)=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)

Vậy a+b+c=14

14

14

\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)

\(=\sqrt{2+5+7+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{5.7}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}=\sqrt{2}+\sqrt{5}+\sqrt{7}\)

\(\Rightarrow a+b+c=2+5+7=14\)

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)

\(a=1+3\sqrt{2}; 3b=30+11\sqrt{8}\Rightarrow \frac{3b}{a}=\frac{30+11\sqrt{8}}{1+3\sqrt{2}}=\frac{(30+11\sqrt{8})(1-3\sqrt{2})}{(1+3\sqrt{2})(1-3\sqrt{2})}\)

\(=\frac{102+68\sqrt{2}}{17}=6+4\sqrt{2}=(2+\sqrt{2})^2\)

\(\Rightarrow \sqrt{\frac{3b}{a}}=2+\sqrt{2}\)

\(\Rightarrow B=\sqrt{2}-(2+\sqrt{2})=-2\)

c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)

f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)

g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)

h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)