`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`

`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`

`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`

`=2sqrt2-8/(2sqrt2)`

`=2sqrt2-sqrt2=sqrt2`

`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`

`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`

`=3-sqrt5-2(sqrt5-1)+3sqrt5`

`=3+3sqrt5-3sqrt5+2=5`

a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)

\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)

\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)

a) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)

\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{2}-2\right)\)

\(=\left|\sqrt{2}+1\right|-\sqrt{2}+2\)

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

\(=3\)

b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)

\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)

\(=-8\sqrt{6}+2\sqrt{6}\)

\(=-6\sqrt{6}\)

c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

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

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

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

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

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

\(=-4\)

a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)

\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)

\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)

\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)

\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)

\(=3\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

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

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

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

\(=\dfrac{3\sqrt{5}}{2}\)

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

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

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

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

\(=\dfrac{1-3}{2}\)

\(=-1\)

\(\dfrac{3\sqrt{18}-2\sqrt{8}}{\sqrt{50}}\)

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

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

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

\(=1\)

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

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

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

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

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

\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

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

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

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

=-2

4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

=4

a. \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

= \(\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\)

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

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

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

b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

= \(\dfrac{\left(\sqrt{15}-\sqrt{5}\right).\left(\sqrt{3}+1\right)}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(2\sqrt{5}+4\right)}{4}\)

=\(\dfrac{\sqrt{45}+\sqrt{15}-\sqrt{15}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).2\left(\sqrt{5}+2\right)}{4}\)

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

= \(\dfrac{2\sqrt{5}}{2}+\dfrac{5\sqrt{5}+10-10-4\sqrt{5}}{2}\)

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

= \(\dfrac{3\sqrt{5}}{2}\)

c. \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}\right):\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)

= \(\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right).\left(\sqrt{5}+\sqrt{2}\right)}.\left(\sqrt{2}+1\right)^2\)

= \(\dfrac{2\sqrt{5}}{3}.\left(2+2\sqrt{2}+1\right)\)

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

= \(\dfrac{6\sqrt{5}+4\sqrt{10}}{3}\)

d. \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)

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

= \(\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)

= \(\left(-2\sqrt{3}-5+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)

= \(\dfrac{-4\sqrt{3}-10+15+5\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}+5}\)

= \(\dfrac{\sqrt{3}+5}{2}.\dfrac{1}{\sqrt{3}+5}\)

= \(\dfrac{1}{2}\)

Nếu đúng cho 1 like nhé!

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

\(=4-1-\dfrac{2}{5}\)

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)

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

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

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

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

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

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

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

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

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

\(B=31\)

_____________________________

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

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

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

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

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