a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)

b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)

c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)

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

\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)

Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3

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

\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)

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

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

\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)

\(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

b. 

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

\(=\sqrt{3-\sqrt{5}}.\sqrt{9-5}+\sqrt{3+\sqrt{5}}.\sqrt{9-5}\)

\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{10}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}\)

\(=\sqrt{10}-\sqrt{2}+\sqrt{10}+\sqrt{2}=2\sqrt{10}\)

c.

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

1)

a)\(\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2=x-y\)

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

c)\(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-\left(\sqrt{5}\right)^2\)

=\(4-5=1\)

d)\(\sqrt{\left(-5\right)^2.3^2}=\left(5\right).3=15\)

e)\(\sqrt{\frac{5}{27}.\frac{8}{20}}=\sqrt{\frac{2}{27}}\)

ĐÂy này nhớ **** vài câu nha

a, \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{2}\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{7-2.\sqrt{3}.\sqrt{7}+3}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\)

a) \(-\sqrt{3}\)      b) -10             c)  60               d)  -1             e) 1

Giúp mình :<

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

\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)

\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)