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a) \(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)\(\Leftrightarrow A=\left[\left(\sqrt{57}+6\right)+\left(3\sqrt{6}+\sqrt{38}\right)\right]\left[\left(\sqrt{57}+6\right)-\left(3\sqrt{6}+\sqrt{38}\right)\right]\)\(\Leftrightarrow A=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)
\(\Leftrightarrow A=57+12\sqrt{57}+36-54-12\sqrt{57}-38\)
\(\Leftrightarrow A=1\)
b) \(B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\sqrt{8+4\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)
c)\(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{3^2-2\times3\times2\sqrt{5}+\left(2\sqrt{5}\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)
\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)
\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)
\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
a)
=\(\sqrt{18-2.3\sqrt{2}.1+1}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1\)
b)
=\(\sqrt{12+2.2\sqrt{3}.3+9}\)
=\(\sqrt{\left(2\sqrt{3}+3\right)^2}\)
=\(2\sqrt{3}+3\)
c)
=\(\sqrt{25-2.5.4\sqrt{2}+32}\)
=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)
=\(4\sqrt{2}-5\)
d)
\(=\sqrt{\left(3-2.\sqrt{3}.\sqrt{2}+2\right)\left(3-2\sqrt{3}+1\right)}\\ =\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}\\ =\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\\ =3-\sqrt{3}-\sqrt{6}+\sqrt{2}\)
e)
\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)
g)
\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
\(x=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}=\dfrac{3}{3}=1\)
\(A=\left(3\cdot1+8\cdot1+2\right)^{2018}=13^{2018}\)
Mẫu của x
\(\sqrt{5}+\sqrt{3^2-2.3.\sqrt{5}+5}=\sqrt{5}+\left|3-\sqrt{5}\right|=3\)
Tử của x
\(\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}=\left(\sqrt{5}+2\right)\sqrt[3]{\left(5\sqrt{5}\right)-3.\left(\sqrt{5}\right)^2.2+3.\sqrt{5}.2^2-2^3}=\left(\sqrt{5}+2\right)\sqrt{\left(\sqrt{5}-2\right)^3}=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=5-4=1\)
=> \(x=\dfrac{1}{3}\)
\(A=\left(\dfrac{3}{3^3}+\dfrac{8}{3^2}+2\right)^{1998}=\left(\dfrac{1+8+9}{3^2}\right)^{1998}=2^{1998}\)
Ta có : \(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)
\(=57+12\sqrt{57}+36-\left(54+12\sqrt{57}+38\right)\)
\(=93-92=1\)
Vậy : \(D=1\)
\(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-2\sqrt{6}-\sqrt{38}+6\right)\)
\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)
\(=\left(93+12\sqrt{57}\right)-\left(92+12\sqrt{57}\right)\)
\(=1\)