1. \(\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\) .
2. \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}.}\)
3. \(\sqrt{11}-\sqrt{20-6\sqrt{11}}=3\)
4.\(\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=2\sqrt{5.}\)
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
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
KT
10 tháng 7 2018
\(\sqrt{\left(2\sqrt{2}-3\right)^2}+2\sqrt{2}=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3\)
\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}=\left|\sqrt{10}-3\right|+\left|\sqrt{10}-4\right|\)
\(=\sqrt{10}-3+4-\sqrt{10}=1\)
\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|2-\sqrt{3}\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
KT
10 tháng 7 2018
\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{\left(6-\sqrt{5}\right)^2}-\sqrt{\left(6+\sqrt{5}\right)^2}\)
\(=6-\sqrt{5}-6-\sqrt{5}=-2\sqrt{5}\)
\(A=\sqrt{49a^2}+3a=7\left|a\right|+3a\)
Nếu \(a\ge0\)thì: \(A=7a+3a=10a\)
Nếu \(a< 0\)thì: \(A=-7a+3a=-4a\)
\(B=3\sqrt{9a^6}-6a^3=9\left|a^3\right|-6a^3\)
Nếu \(a\ge0\)thì: \(B=9a^3-6a^3=3a^3\)
Nếu \(a< 0\)thì: \(B=-9a^3-6a^3=-15a^3\)
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,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
29 tháng 9 2020
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
29 tháng 9 2020
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
20 tháng 8 2021
a: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
\(1.\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}=5+\sqrt{7}-\sqrt{7-2\sqrt{7}+1}=5+\sqrt{7}-\sqrt{7}+1=6\)
\(2.\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\sqrt{3}+1=2\)
\(3.VT=\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{11-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{11}+3=3=VP\)
Vậy , đẳng thức được chứng minh .
\(4.VT=\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=\sqrt{36+2.6\sqrt{5}+5}-\sqrt{36-2.6\sqrt{5}+5}=6+\sqrt{5}-6+\sqrt{5}=2\sqrt{5}=VP\)
Vậy , đẳng thức được chứng minh .