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những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)
\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)
\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)
\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
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)
\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
\(1.\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\)
\(2.\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)
\(3.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}-\sqrt{3}\)
\(4.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(5.\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{4+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
\(6.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c)
\(\sqrt{2}C=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)
\(=\sqrt{5}+1-\left(\sqrt{5}-1\right)-2=0\Rightarrow C=0\)
b)
\(B=3\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)\)
\(\Rightarrow\sqrt{2}B=3\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\)
\(=3\left(\sqrt{5}+1+\sqrt{5}-1\right)-\sqrt{5}\left(\sqrt{5}+1-\sqrt{5}+1\right)\)
\(\sqrt{2}B=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\Rightarrow B=2\sqrt{10}\)
C)√3+√5−√3−√5−√2b) (3−√5)√3+√5+(3+√5)√3−√5d) √4−√7−√4+√7+√7e) √6,5+√12+√6,5−√12+2√6mình cần giải gấp ạ
+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)
\(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)
\(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)
\(=4\sqrt{3}\approx6,9282\)
+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)
\(=\sqrt{x-9+6\sqrt{x-9}+9}\)
\(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)
\(=\left|\sqrt{x-9}-3\right|\)
\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)
\(a,\dfrac{-3}{5}.\sqrt{\left(-0.5\right)^2}\\ =\dfrac{-3}{5}.0,5\\ =\dfrac{-3}{5}.\dfrac{1}{2}\\ =-\dfrac{3}{10}\)
Câu (b) nhìn hơi lạ lạ á :v
\(c,\sqrt{\left(1-\sqrt{7}\right)^2}+\sqrt{7}\\ =\sqrt{7}-1+\sqrt{7}\\ =2\sqrt{7}-1\)
\(d,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =3+\sqrt{2}-\left(3-\sqrt{2}\right)\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)
a) \(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
b) \(\sqrt{13-4\sqrt{3}}=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+1}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}=2\sqrt{3}-1\)
c) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
d) \(\sqrt{3+2\sqrt{2}+\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{3+2\sqrt{2}+\sqrt{\left(2-\sqrt{2}\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}+2-\sqrt{2}}\)
\(=\sqrt{5+\sqrt{2}}\)
e) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
\(=2+\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)
\(=2+\sqrt{9-4\sqrt{5}}\)
\(=2+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2+\sqrt{5}-2=\sqrt{5}\)
f) đề sai nhé:
\(\sqrt{3a}.\sqrt{12a}=\sqrt{36a^2}=6a\)\(\left(a\ge0\right)\)
g) \(\sqrt{16a^2b^8}=4b^4\left|a\right|\)
h) \(\sqrt{7a}.\sqrt{63a^3}=\sqrt{441.a^4}=21a^2\)
a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)
c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)
d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)
e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)
f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)