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15 tháng 2 2020

a)

\((\sqrt2- \sqrt3).(\sqrt2+\sqrt3)\)

=\(\sqrt2.\sqrt2 + \sqrt2.\sqrt3-\sqrt3.\sqrt2+\sqrt3.\sqrt3\)

=\(1.1+1.\sqrt3-\sqrt3.1+\sqrt3.\sqrt3\)

=1+0+3=4

15 tháng 2 2020

\(a,\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)=\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2=2-3=-1\)

\(b,-\left(\sqrt{2}\right)^4+\left(\sqrt{3}\right)^6=-\left(\sqrt{2}^2\right)^2+\left(\sqrt{3}^2\right)^3=-2^2+3^3=-4+27=23\)

\(c,A=\frac{1}{1-\frac{1}{1-2^{-4}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}=\frac{1}{1-\frac{1}{1-\frac{1}{16}}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}=\frac{1}{1-\frac{1}{\frac{15}{16}}}+\frac{1}{1+\frac{1}{\frac{3}{2}}}\)

\(=\frac{1}{1-\frac{16}{15}}+\frac{1}{1+\frac{2}{3}}=\frac{1}{-\frac{1}{15}}+\frac{1}{\frac{5}{3}}=-15+\frac{3}{5}=-14,4\)

\(d,B=9+99+...+99...9=\left(10-1\right)+\left(100-1\right)+...+\left(100...0-1\right)\)

\(=\left(10+100+...+100...0\right)-\left(1+1+...+1\right)=11...10-50=11...1060\)(có 48 chữ số 1)

12 tháng 8 2019

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

12 tháng 8 2019

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

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

16 tháng 7 2018

a

27 tháng 8 2019

a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)

b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)

c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)

d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)

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

f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)

g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)

17 tháng 7 2019

a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)

\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)

\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)

b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)

\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)

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

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

\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)

c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)

\(=27+12\sqrt{5}+12\sqrt{5}\)

\(=27+24\sqrt{5}\)

d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)

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

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

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

= 8

f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)

\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

= 14

17 tháng 7 2019

a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)

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

= 9 (đpcm)

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

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

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

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)

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

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

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

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

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

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)

\(=2\sqrt{3}+\sqrt{6}\)

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)

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

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

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)