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\(a.\left(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}-\dfrac{2-\sqrt{5}}{2+\sqrt{5}}\right).\dfrac{\sqrt{2}}{23}=\dfrac{\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}.\dfrac{\sqrt{2}}{23}=-8\sqrt{5}.\dfrac{\sqrt{2}}{23}\)
\(b.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\)
a: \(A=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}-\sqrt{2}=0\)
b: \(B=\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)=1-2=-1\)
c: \(B=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=-\sqrt{6}+2\)
a: \(=\sqrt{15}-3+\sqrt{15}=2\sqrt{15}-3\)
b: \(=\left(\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}\right)-2\sqrt{3}+2\)
\(=2\)
c: \(=\left(\sqrt{3}-\sqrt{2}\right)\cdot3\cdot\left(\sqrt{3}+\sqrt{2}\right)=3\)
B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(a.\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\dfrac{\left(2+\sqrt{3}\right)\sqrt{3-2\sqrt{3}+1}}{\sqrt{3+2\sqrt{3}+1}}=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{2\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{3-1}=4-3=1\)
\(b.\dfrac{\left(\sqrt{5}-1\right)^3}{\sqrt{5}-2}=\dfrac{5\sqrt{5}-15+3\sqrt{5}-1}{\sqrt{5}-2}=\dfrac{8\sqrt{5}-16}{\sqrt{5}-2}=\dfrac{8\left(\sqrt{5}-2\right)}{\sqrt{5}-2}=8\)
\(c.\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left[\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right]=2\left(3+1+3\right)=2.7=14\)
\(d.\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=\dfrac{2}{2}=1\)
Bài 2:
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(P=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2.\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(P=\left[\dfrac{\left(a-1\right)^2}{4a}\right].\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\sqrt{a}-1}{a-1}\right)\)
\(P=\dfrac{\left(a-1\right)^2}{4a}.\dfrac{2\sqrt{a}.\left(-2\right)}{a-1}\)
\(P=\dfrac{\left(a-1\right)^2\left(-4\sqrt{a}\right)}{4a.\left(a-1\right)}\)
\(P=\dfrac{\left(a-1\right).\left(-\sqrt{a}\right)}{a}=\dfrac{-a\sqrt{a}+\sqrt{a}}{a}\)
Bài 1:
\(A=\dfrac{2}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}\)\(A=\dfrac{2\sqrt{2}}{2}-\dfrac{1\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2-1}\)
\(A=\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{1}+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(A=\sqrt{2}-\sqrt{3}-\sqrt{2}+\sqrt{3}+1\)
\(A=1\)
\(a.\left(\dfrac{\sqrt{6}-\sqrt{3}}{5\sqrt{2}-5}+\dfrac{\sqrt{5}}{5}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}=\left[\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right]:\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{\sqrt{3}+\sqrt{5}}{5}.\dfrac{1}{\sqrt{5}+\sqrt{3}}=\dfrac{1}{5}\)
\(b.\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=\dfrac{-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=2\sqrt{3}-\dfrac{4}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}+1\right)-4}{\sqrt{3}+1}=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\left(1+\sqrt{3}\right)}{1+\sqrt{3}}=2\)
Câu a : \(\left(\sqrt{80}+\sqrt{20}\right):\sqrt{45}=\sqrt{80}:\sqrt{45}+\sqrt{20}:\sqrt{45}=\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{4}{9}}=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
Câu b : \(\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{18}+\sqrt{27}\right)=\sqrt{54}+\sqrt{81}-\sqrt{36}-\sqrt{54}=\sqrt{81}-\sqrt{36}=9-6=3\)
Câu c : \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}-\dfrac{6}{\sqrt{15+3}}=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}-\dfrac{6}{\sqrt{18}}\)
\(=\sqrt{15}-\dfrac{6}{\sqrt{18}}=\dfrac{\sqrt{270}-6}{3\sqrt{2}}=\dfrac{3\sqrt{30}-6}{3\sqrt{2}}=\dfrac{3\left(\sqrt{30}-6\right)}{3\sqrt{2}}=\dfrac{\sqrt{30}-2}{\sqrt{2}}=\sqrt{15}-\sqrt{2}\)
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4