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♉ⓃⒶⓂ๖P๖S๖Pツ
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Nguyễn Ngọc Lộc
19 tháng 7 2020 lúc 18:31

Ta có : \(\frac{20\sqrt{300}-15\sqrt{675}+5\sqrt{75}}{\sqrt{15}}=\frac{\sqrt{15}\left(20\sqrt{20}-15\sqrt{45}+5\sqrt{5}\right)}{\sqrt{15}}\)

\(=20\sqrt{20}-15\sqrt{45}+5\sqrt{5}=40\sqrt{5}-45\sqrt{5}+5\sqrt{5}=0\)

Phạm văn đồng
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Khánh Trang Lê
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Hoang Minh
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Nguyễn Lê Phước Thịnh
23 tháng 6 2023 lúc 19:58

k: =3căn 2-2căn 3+2căn 3-2=3căn2-2

l: =20*căn 20-15*căn 45+5*căn 5

=40căn 5-45căn 5+5căn 5=0

Khánh Trang Lê
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* Nhók EXO - L dễ thưng...
5 tháng 10 2017 lúc 21:16

a,

\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)

\(=6\sqrt{6}^2-12\sqrt{9.2}+15\sqrt{4.3}-\dfrac{3}{4}\sqrt{16.3}\)

= \(6.6-12.3\sqrt{2}+15.2\sqrt{3}-\dfrac{3}{4}.4\sqrt{3}\)

\(=36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\)

\(=36-36\sqrt{2}+27\sqrt{3}\)

阮芳草
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Minh Nguyễn Cao
4 tháng 8 2018 lúc 19:42

\(\left(20.\sqrt{0.03}+12.\sqrt{3}-\frac{1}{5}.\sqrt{75}\right).\sqrt{6}\)

\(=20.\sqrt{0,03.6}+12.\sqrt{3.6}-\frac{1}{5}.\sqrt{75.6}\)

\(=20.\sqrt{\frac{9}{50}}+12.\sqrt{3^2.2}-\frac{1}{5}.\sqrt{15^2.2}\)

\(=6\sqrt{2}+36\sqrt{2}-3\sqrt{2}\)

\(=39\sqrt{2}\)

Ngọc Châu
4 tháng 8 2018 lúc 19:49

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

\(=\sqrt{6}.13\sqrt{3}=13\sqrt{18}=39\sqrt{2}\)

Phạm thị thảo ngân
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Mới vô
13 tháng 7 2017 lúc 7:53

a,

\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\\ =\sqrt{3-2\cdot1\cdot\sqrt{3}+1}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot1\cdot\sqrt{3}+1^2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}\\ =-1\)

b,

\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-3+\sqrt{2}\\ =\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)

c,

\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}\\ =\sqrt{5+2\cdot\sqrt{2\cdot5}+2}-\sqrt{5-2\cdot\sqrt{2\cdot5}+2}\\ =\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}\\ =2\sqrt{2}\)

d,

\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\\ =\left(20\cdot\sqrt{20}\cdot\sqrt{15}-15\cdot\sqrt{45}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\left(20\cdot2\cdot\sqrt{5}\cdot\sqrt{15}-15\cdot3\cdot\sqrt{5}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\sqrt{15}\cdot\left(20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\right):\sqrt{15}\\ =20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\\ =40\sqrt{5}-45\sqrt{5}+5\sqrt{5}\\ =0\)

Phạm Hà Linh
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HT.Phong (9A5)
8 tháng 9 2023 lúc 5:59

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

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

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

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

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

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

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

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

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

ngan kim
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Hquynh
18 tháng 8 2023 lúc 19:19

\(=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{\sqrt{2}.\sqrt{2}.\sqrt{5}}{\sqrt{2}.\sqrt{3}}\\ =\dfrac{\sqrt{10}}{\sqrt{3}}=\dfrac{\sqrt{30}}{3}\)

HaNa
18 tháng 8 2023 lúc 19:19

\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\\ =\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =2\sqrt{5}:\sqrt{6}\\ =2\sqrt{5}.\dfrac{1}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}.\sqrt{6}}{6}\\ =\dfrac{1}{3}.\sqrt{30}\\ =\dfrac{\sqrt{30}}{3}\)