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Byun Baekhyun
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Bình Lê
10 tháng 7 2017 lúc 15:39

\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)

Bình Lê
10 tháng 7 2017 lúc 15:45

\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)

Bình Lê
10 tháng 7 2017 lúc 15:48

\(C=\dfrac{1}{2}\left(11+2\sqrt{30}\right)-\dfrac{\sqrt{30}}{2}-\dfrac{\sqrt{30}}{2}\\ =\dfrac{11}{2}+\sqrt{30}-\sqrt{30}\\ =\dfrac{11}{2}\)

Như Huỳnh
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qwerty
19 tháng 6 2017 lúc 10:34

a) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=1\)

qwerty
19 tháng 6 2017 lúc 11:38

b)

\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)

\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2\sqrt{6}}{6}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)

\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)

\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}\)

\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{6+3\sqrt{3}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}}{6\sqrt{3}}}\)

\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}\)

\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

\(=\dfrac{\sqrt{6+3\sqrt{3}}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\left(-\sqrt{3}+2+\sqrt{3}\right)}}{-2\sqrt{3}}\)

\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}\)

\(=\dfrac{\sqrt{12+6\sqrt{3}}}{-2\sqrt{3}}\)

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

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

\(=-\dfrac{\left(3+\sqrt{3}\right)\sqrt{3}}{6}\)

\(=-\dfrac{3\sqrt{3}+3}{6}\)

\(=-\dfrac{3\left(\sqrt{3}+3\right)}{6}\)

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

qwerty
19 tháng 6 2017 lúc 11:58

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

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

\(=\dfrac{1-\sqrt{1+\dfrac{\sqrt{3}}{2}}+\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3\left(1+\dfrac{\sqrt{3}}{2}\right)}}{2}}{-\dfrac{\sqrt{3}}{2}}+\dfrac{\left(1-\dfrac{\sqrt{3}}{2}\right)\cdot\left(1+\sqrt{1-\dfrac{\sqrt{3}}{2}}\right)\cdot2}{\sqrt{3}}\)

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

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

\(=\dfrac{-\left(2-2\sqrt{1+\dfrac{\sqrt{3}}{2}}+\sqrt{3}-\sqrt{3+\dfrac{3\sqrt{3}}{2}}\right)+2\cdot2\sqrt{1-\dfrac{\sqrt{3}}{2}}-\sqrt{3}-\sqrt{3-\dfrac{3\sqrt{2}}{2}}}{\sqrt{3}}\)

\(=1\)

Hương Giang
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Chí Cường
8 tháng 10 2017 lúc 16:22

a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)

nguyen ngoc son
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MC Alextion
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Phùng Khánh Linh
2 tháng 10 2018 lúc 18:09

\(VT=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{48}+\sqrt{49}}=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+\dfrac{\sqrt{4}-\sqrt{3}}{1}+...+\dfrac{\sqrt{49}-\sqrt{48}}{1}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{49}-\sqrt{48}=\sqrt{49}-1=7-1=6=VP\)Vậy , đẳng thức được CM

Trang Nguyễn
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Lê Thị Thục Hiền
24 tháng 8 2021 lúc 6:13

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

HoàngIsChill
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Nguyễn Huy Tú
15 tháng 7 2021 lúc 14:46

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Phạm Hà Linh
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Nguyễn Lê Phước Thịnh
5 tháng 9 2023 lúc 14:32

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

=>\(\dfrac{3}{2}\sqrt{x-3}=3\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

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

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

Nguyen Ngoc Lien
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Nguyễn Lê Phước Thịnh
3 tháng 9 2022 lúc 23:03

a: \(=\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\cdot\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\)

\(=5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)

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

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

\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

d: \(=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)