Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

=5

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(\sqrt{3}+2\right)}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

= \(\sqrt{4+\sqrt{5\left(\sqrt{3}+5-\sqrt{3}\right)}}\)

= \(\sqrt{4+\sqrt{25}}\)

= \(\sqrt{4+5}=3\)

b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)

\(a,\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{2}\)

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

\(b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{25}=5\)

a)=1

b)=25

\(A=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

     \(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\times2\times\sqrt{3}}}}\)

     \(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)

     \(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

     \(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

     \(=\sqrt{5\sqrt{3}+5\sqrt{25+3-2\times5\times\sqrt{3}}}\)

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

     \(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)             (do  \(5-\sqrt{3}>0\))

     \(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

     \( = {\sqrt{25}} = |5| \)

a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)

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

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

\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}.\)

\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3+4+2\sqrt{12}}}}\)

\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}\)

\(\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

\(\sqrt{5\sqrt{3}+5\sqrt{25+3-2.\sqrt{25.3}}}\)

\(\sqrt{5\sqrt{3}+5\sqrt{\left(\sqrt{25}-\sqrt{3}\right)^2}}\)

\(\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

\(\sqrt{25}=5\)

cho mình hỏi tại sao  10\(\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}\)lại bằng  10\(\sqrt{3}\)

là tại vì \(10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}..\)

\(10\left(\sqrt{3}+2\right)=10\sqrt{3}+20\)

 \(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)

\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)

Vậy...

~ Chắc chắn đúng cậu nhé ~ Tiếc gì 1 tk cho tớ nào?

Bài 1:

a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)

\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)

\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)

\(=\dfrac{5}{2}\sqrt{5}\)

c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)

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

Bài 2:

e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)

Ta có: \(\sqrt{6-x}=3x-4\)

\(\Leftrightarrow6-x=\left(3x-4\right)^2\)

\(\Leftrightarrow9x^2-24x+16+6-x=0\)

\(\Leftrightarrow9x^2-25x+22=0\)

\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)

Vậy: Phương trình vô nghiệm