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Nguyên Phan
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Nguyễn Lê Phước Thịnh
12 tháng 7 2021 lúc 1:09

2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)

\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)

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

\(=-11\)

3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)

\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)

\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)

\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)

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Qúy Công Tử
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Nguyễn Lê Phước Thịnh
9 tháng 8 2022 lúc 13:45

1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)

2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)

3: \(=\sqrt{3}+1-\sqrt{3}=1\)

 

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phan anh thư
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Thư Thư
23 tháng 6 2023 lúc 9:05

\(a,\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{2}\right)}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-3\sqrt{2}}{4-6}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{2}.\sqrt{3}}.\dfrac{4\sqrt{3}}{-2}-\dfrac{1}{\sqrt{2}}\)

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

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

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

\(=\dfrac{\sqrt{2}+\sqrt{3}-1+2+\sqrt{6}-\sqrt{6}-3-\sqrt{2}-\sqrt{3}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)

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

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

 

 

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Ngô Hiểu Phong
23 tháng 6 2023 lúc 9:04

.

 

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Như Huỳnh
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qwerty
19 tháng 6 2017 lúc 10:34

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

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

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

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

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

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

\(=1\)

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qwerty
19 tháng 6 2017 lúc 11:38

b)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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qwerty
19 tháng 6 2017 lúc 11:58

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

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

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

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

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

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

\(=1\)

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griselda
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Nguyễn Lê Phước Thịnh
22 tháng 8 2021 lúc 23:40

\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)

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

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An Đinh Khánh
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Thanh Phong (9A5)
21 tháng 7 2023 lúc 10:57

a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

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

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

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

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

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

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

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

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Nguyễn Lê Phước Thịnh
21 tháng 7 2023 lúc 10:58

a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)

b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

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manh
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Thanh Phong (9A5)
30 tháng 9 2023 lúc 6:16

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

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

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

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

\(=2\sqrt{3}\)

b) \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)

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

\(=\sqrt{2}-\sqrt{2}\)

\(=0\)

c) \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\cdot\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)

\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)

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

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

\(=9-4\sqrt{5}\)

d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(=6-121\)

\(=-115\)

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Trang Nguyễn
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Yeutoanhoc
28 tháng 6 2021 lúc 8:39

Bạn chia nhỏ ra để nhận được câu tl sớm nhất nhé!Bạn đặt câu hỏi free mà để dày cộp như này khum ai dám làm =(((

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

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

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