Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)

a) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=\left(2+6+15-36\right)\sqrt{3}=-13\sqrt{3}\)
b) \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=6\left(3+8-5\right)=36\)
a)\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=\sqrt{4\cdot3}+2\sqrt{9\cdot3}+3\sqrt{25\cdot3}-9\sqrt{16\cdot3}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
b)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)

a/ Đề sai
b/ \(\sqrt{125}-4\sqrt{45}+3\sqrt{2}-\sqrt{80}=5\sqrt{5}-12\sqrt{5}+3\sqrt{2}-4\sqrt{5}\)
\(=-11\sqrt{5}+3\sqrt{2}\)
c/ \(2\sqrt{\frac{27}{4}}-\sqrt{\frac{48}{9}}-\frac{2}{5}\sqrt{\frac{75}{16}}=2.\frac{3\sqrt{3}}{2}-\frac{4\sqrt{3}}{3}-\frac{2}{5}.\frac{5\sqrt{3}}{4}\)
\(=3\sqrt{3}-\frac{4\sqrt{3}}{3}-\frac{\sqrt{3}}{2}=\sqrt{3}\left(3-\frac{4}{3}-\frac{1}{2}\right)=\frac{7\sqrt{3}}{6}\)
d/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)

a)\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
b) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}=4\sqrt{3}\)
c)\(\sqrt{12}+2\sqrt{75}-3\sqrt{48}-\frac{2}{7}\sqrt{147}=2\sqrt{3}+10\sqrt{3}-12\sqrt{3}-2\sqrt{3}=-2\sqrt{3}\)
d) \(\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{9-4\sqrt{5}}\)
\(=\left|3+\sqrt{5}\right|-\sqrt{\left(\sqrt{5}-2\right)^2}=3+\sqrt{5}-\left|\sqrt{5}-2\right|=3+\sqrt{5}-\sqrt{5}+2=5\)
e) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{\sqrt{5}+\sqrt{2}}{3}\)
\(=\left[\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}=-3\)
Nản k lm nữa ^^

a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)

b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)

1) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}-10-5\sqrt{10}\)
\(=-10\)
2) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=14-2\sqrt{21}-7+2\sqrt{21}\)
\(=7\)
3) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\) (hẳn đề là như thế này)
\(=33-3\sqrt{22}-11+3\sqrt{22}\)
\(=22\)

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) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
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
\(A=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)
\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=2\sqrt{16.5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
\(B=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)
\(=2\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}+3\sqrt{22}=22\)