6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)

7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)

9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

a: \(=3\sqrt{5}+2\sqrt{5}-2\sqrt{5}=3\sqrt{5}\)

b: \(=2\sqrt{2}+2\sqrt{2}+5\sqrt{2}=9\sqrt{2}\)

c: \(=4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+2\sqrt{5}=7\sqrt{3}-\sqrt{5}\)

d: \(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\)

e: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)

=7-2*căn 21+2*căn 21

=7

f: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}\)

=22-3*căn 22+3*căn 22

=22

a) \(3\sqrt{5}+\sqrt{20}-2\sqrt{5}\)

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

\(=3\sqrt{5}\)

b) \(2\sqrt{2}+\sqrt{8}+\sqrt{50}\)

\(=2\sqrt{2}+2\sqrt{2}+5\sqrt{2}\)

\(=9\sqrt{5}\)

c) \(4\sqrt{3}+\sqrt{27}-\sqrt{45}+2\sqrt{5}\)

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

\(=7\sqrt{3}-\sqrt{5}\)

d) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)

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

\(=-\sqrt{3}\)

e) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}\)

\(=7\)

f) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)

\(=22-3\sqrt{22}+3\sqrt{22}\)

\(=22\)

g) \(3\sqrt{45}-5\sqrt{125x}+7\sqrt{20x}+28\)

\(=9\sqrt{5}-25\sqrt{5x}+14\sqrt{5x}+28\)

\(=9\sqrt{5}-11\sqrt{5x}+28\)

\(\left(\sqrt{12}+\sqrt{27}-\sqrt{18}\right)\cdot3\\ =(\sqrt{4\cdot3}+\sqrt{9\cdot3}-\sqrt{6}\cdot\sqrt{3})\cdot\sqrt{3}\\ =\left(2\sqrt{3}+3\sqrt{3}-\sqrt{6}\cdot\sqrt{3}\right)\cdot\sqrt{3}\\ =2\cdot3+3\cdot3-\sqrt{6}\cdot3\\ =6+9-3\sqrt{6}\\ =15-3\sqrt{6}\)

\(\left(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\right):\sqrt{5}\\ =\left(15\sqrt{4\cdot5}-3\sqrt{9\cdot5}+2\sqrt{5}\right):\sqrt{5}\\ =\left(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\right):\sqrt{5}\\ =30-9+2\\ =23\)

Tính:

\(H=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\dfrac{3\sqrt{4.2}-2\sqrt{4.3}+\sqrt{4.5}}{3\sqrt{9.2}-2\sqrt{9.3}+\sqrt{9.5}}\)

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

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

\(=\dfrac{2}{3}\)

Cảm ơn bạn !

\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)

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

___________

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

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

__________

\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)

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

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

\(=\dfrac{2}{3}\)

a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)

b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

kết quả hay cả lời giải

ILoveMath                                                          , lời giải

a) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}=\sqrt{2^2\cdot3}+5\sqrt{3}-\sqrt{4^2\cdot3}\)

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

b) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}=5\sqrt{5}+\sqrt{2^2\cdot5}-3\sqrt{3^2\cdot5}\) \(=5\sqrt{5}+2\sqrt{5}-9\sqrt{5}=-2\sqrt{5}\)

c)

\(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}=2\sqrt{4^2\cdot2}+4\sqrt{2^2\cdot2}-5\sqrt{3^2\cdot2}\) \(=8\sqrt{2}+8\sqrt{2}-15\sqrt{2}=\sqrt{2}\)

d)\(\sqrt{2^2\cdot3}+\sqrt{5^2\cdot3}-\sqrt{3^2\cdot3}=2\sqrt{3}+5\sqrt{3}-3\sqrt{3}=4\sqrt{3}\)

Lời giải:

\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)

\(=\sqrt{4+\sqrt{8}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}\)

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

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

\(y=\frac{3.2\sqrt{2}-2.2\sqrt{3}+2\sqrt{5}}{3.3\sqrt{2}-2.3\sqrt{3}+3\sqrt{5}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)

\(=\frac{2(3\sqrt{2}-2\sqrt{3}+\sqrt{5})}{3(3\sqrt{2}-2\sqrt{3}+\sqrt{5})}=\frac{2}{3}\)

Ta có: \(b=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

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

\(=\dfrac{2}{3}\)

Ta có: \(a=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)

\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)

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

=2

Thay a=2 và \(b=\dfrac{2}{3}\) vào M, ta được:

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

\(=\dfrac{7}{8}+\dfrac{1}{4}\)

\(=\dfrac{7}{8}+\dfrac{2}{8}=\dfrac{9}{8}\)