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\(a.\dfrac{\sqrt{8}+2}{\sqrt{2}+1}.\sqrt{2}=\dfrac{\sqrt{2}.2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\sqrt{2}\)
\(b.\dfrac{5+\sqrt{5}}{\sqrt{5}+1}.\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}.\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\sqrt{5}.\sqrt{5}=5\)
\(c.\dfrac{2}{\sqrt{2}}+\sqrt{18}+\sqrt{32}=\sqrt{2}+\sqrt{9.2}+\sqrt{16.2}=\sqrt{2}+3\sqrt{2}+4\sqrt{2}=8\sqrt{2}\)
\(d.\left(1-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(1+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\right)=\left(1-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(1+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\right)=\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)=1-2=-1\)
a: \(B=\dfrac{12+2\sqrt{35}+12-2\sqrt{35}}{2}-\dfrac{5}{2}\)
=12-5/2
=9,5
b: \(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+4-2\sqrt{5}+4}{1}=8\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
a) \(2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}=2\sqrt{2^2.5}-\sqrt{5^2.2}+3\sqrt{4^2.5}-\sqrt{8^2.5}\\ =4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)
b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=\sqrt{4^2.2}-\sqrt{5^2.2}+\sqrt{3^2.2}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
c) \(3\sqrt{3}+4\sqrt{2}-5\sqrt{27}=3\sqrt{3}+4\sqrt{2}-5\sqrt{3^2.3}=3\sqrt{3}+4\sqrt{2}-15\sqrt{3}=4\sqrt{2}-12\sqrt{3}\)
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}\\ =\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\left(\sqrt{3+1}\right)^2-1^2}\\ =\dfrac{2\sqrt{3}}{\sqrt{3}}=2\)
e)\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=2^2-\left(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)^2=4-\left(\dfrac{9+6\sqrt{3}+3}{3+2\sqrt{3}+1}\right)\\ =4-\left(\dfrac{6\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}\right)=4-3=1\)
b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=\left(4-5+3\right)\sqrt{2}=2\sqrt{2}\)
a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)
b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)
c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)
d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)
f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)
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: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
a: \(\dfrac{\sqrt{50}-\sqrt{32}+\sqrt{8}}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}-4\sqrt{2}+2\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{\sqrt{2}}=3\)
b: \(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
\(=\dfrac{4\left(\sqrt{5}+1\right)}{5-1}-\sqrt{5}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1