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

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

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

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

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

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

=\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{4+2.2\sqrt{3+\left(\sqrt{3}\right)^2}}}}\)

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

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

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

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

=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}\right)^2}+2.5.\sqrt{3}+5^2}}\)

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

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

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

=\(10\sqrt{3+10}\)

=\(\sqrt{10\left(\sqrt{3+1}\right)}\)

17 tháng 8 2020

Bài làm:

a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)

\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)

\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)

\(A=4+2\sqrt{3}+5\sqrt{3}-1\)

\(A=3+7\sqrt{3}\)

b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)

\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)

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

\(A=2\)

17 tháng 8 2020

Phần b mình viết nhầm tên thành A, bn sửa thành B nhé

c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

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

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

\(C=-3\)

26 tháng 7 2018

E = \(6x+\sqrt{9x^2-12x+4}\)

E = \(6x+\sqrt{\left(3x-2\right)^2}\)

E = \(6x+\left|3x-2\right|\)

E = \(6x+3x-2\)

E = \(9x-2\)

F = \(5x-\sqrt{x^2+4x+4}\)

F = \(5x-\sqrt{\left(x+2\right)^2}\)

F = \(5x-\left|x+2\right|\)

F = \(5x-x+2\)

F = \(4x+2\)

30 tháng 8 2020

Bài làm:

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

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

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

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

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

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

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

Đến đây chịu-.-

24 tháng 5 2017

Hỏi đáp Toán

12 tháng 7 2017

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

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

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

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

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

= 5

\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)

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

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

\(=3\)

a: \(=6\sqrt{2}-12\sqrt{3}-10\sqrt{2}+12\sqrt{3}=-4\sqrt{2}\)

b: \(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)