Rút gọn:
1) \(\left(\sqrt{12}+2\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
2) \(\left(4\sqrt{2}-\sqrt{8}+2\right).\sqrt{2-\sqrt{8}}\)
3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\dfrac{3}{2}\sqrt{12}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 9 2017
\(\left(\sqrt{12}+2\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
\(=\sqrt{12}:\sqrt{3}+2\sqrt{27}:\sqrt{3}-\sqrt{3}:\sqrt{3}\)
\(=\sqrt{4}+2\sqrt{9}-1\)
\(=2+6-1\)
\(=7\)
21 tháng 9 2017
2) \(\left(4\sqrt{2}-\sqrt{8}+2\right).\sqrt{2-\sqrt{8}}\)
\(=\left(4\sqrt{2}-2\sqrt{2}+2\right).\sqrt{2-2\sqrt{2}}\)
\(=\left(2\sqrt{2}+2\right)^2.\left(\sqrt{2-2\sqrt{2}}\right)^2\)
\(=\left(8+4\right)\left(2-2\sqrt{2}\right)\)
\(=12.\left(2-2\sqrt{2}\right)\)
\(=24-24\sqrt{2}\)
\(=24\left(1-\sqrt{2}\right)\)
3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\frac{3}{2}\sqrt{12}\right)\)
\(=\sqrt{3}\left(2\sqrt{3^2.3}-\sqrt{5^2.3}+\frac{3}{2}\sqrt{2^2.3}\right)\)
\(=\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)
\(=\sqrt{3}.4\sqrt{3}\)
\(=12\)
2 tháng 7 2021
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
2 tháng 7 2021
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
MA
3 tháng 1 2019
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 6 2021
\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)
\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)
\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)
\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)
28 tháng 6 2021
a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
23 tháng 9 2021
\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)
\(=9\sqrt{5}-28\sqrt{3}\)
\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(=4-2\cdot5+6-7\)
\(=4-10+6-7\)
=-7
23 tháng 9 2021
A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)
=\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)
=2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)
=\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)
=9\(\sqrt{5}\)- 28\(\sqrt{3}\)
17 tháng 11 2018
a) \(\sqrt{243}-\dfrac{1}{2}\sqrt{12}-2\sqrt{75}+2\sqrt{27}=\sqrt{81.3}-\dfrac{1}{2}.2\sqrt{3}-2\sqrt{25.3}+2\sqrt{9.3}=\sqrt{81}.\sqrt{3}-\sqrt{3}-2\sqrt{25}.\sqrt{3}+2\sqrt{9}.\sqrt{3}=9\sqrt{3}-\sqrt{3}-10\sqrt{3}+6\sqrt{3}=\sqrt{3}\left(9-1-10+6\right)=4\sqrt{3}\)
b) \(\left(2+\sqrt{6}\right)\sqrt{7-4\sqrt{3}}=\left(2+\sqrt{6}\right)\sqrt{4-2\sqrt{3}.2+3}=\left(2+\sqrt{6}\right)\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{6}\right)\left|2-\sqrt{3}\right|=\left(2+\sqrt{6}\right)\left(2-\sqrt{3}\right)=4-2\sqrt{3}+2\sqrt{6}-3\sqrt{2}\)
c) \(\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.\left(3\sqrt{2}+\sqrt{14}\right)=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\left(3\sqrt{2}+\sqrt{14}\right)=\sqrt{\dfrac{1}{8+3\sqrt{7}}}.\left(3\sqrt{2}+\sqrt{14}\right)=\sqrt{\dfrac{8-3\sqrt{7}}{8^2-\left(3\sqrt{7}\right)^2}}.\sqrt{2}.\left(3+\sqrt{7}\right)=\sqrt{\dfrac{2\left(8-3\sqrt{7}\right)}{64-63}}\left(3+\sqrt{7}\right)=\sqrt{16-6\sqrt{7}}.\left(3+\sqrt{7}\right)=\sqrt{9-2.3.\sqrt{7}+7}.\left(3+\sqrt{7}\right)=\sqrt{\left(3-\sqrt{7}\right)^2}.\left(3+\sqrt{7}\right)=\left|3-\sqrt{7}\right|\left(3+\sqrt{7}\right)=\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)=9-7=2\)
3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\dfrac{3}{2}\sqrt{12}\right)\)
=\(\sqrt{3}\left(2.3\sqrt{3}-5\sqrt{3}+\dfrac{3}{2}.2\sqrt{3}\right)\)
=\(\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)
=\(\sqrt{3}.\left(4\sqrt{3}\right)\)
=12
\(\left(2\sqrt{3}+2.3\sqrt{3}-\sqrt{3}\right):\sqrt{3}\)
=\(\left(2\sqrt{3}+6\sqrt{3}-\sqrt{3}\right):\sqrt{3}\)
=\(\left(7\sqrt{3}\right):\sqrt{3}\)
=\(7\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
=7