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\(a.\sqrt{14+4\sqrt{3}.\sqrt{2}}=\sqrt{12+2.2\sqrt{3}.\sqrt{2}+2}=2\sqrt{3}+\sqrt{2}\)
\(b.\sqrt{11-4\sqrt{3}.\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.\sqrt{3}+3}=2\sqrt{2}-\sqrt{3}\)
\(c.\sqrt{28+16\sqrt{3}}=\sqrt{16+2.2\sqrt{3}.4+12}=4+2\sqrt{3}\)
\(d.\sqrt{11+4\sqrt{7}}=\sqrt{7+2.2\sqrt{7}+4}=\sqrt{7}+2\)
\(e.\sqrt{29-4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}+1}=2\sqrt{7}-1\)
\(f.\sqrt{21+6\sqrt{2}.\sqrt{3}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}\)
a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)
mình làm mẫu 2 bài nhé 2 bài kia bạn làm tương tự
1)a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)
\(\sqrt{10-2\sqrt{21}}+\sqrt{7}=\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}+\sqrt{7}=\sqrt{7}+\sqrt{3}+\sqrt{7}=2\sqrt{7}+\sqrt{3}\)
2)a) \(\sqrt{12-6\sqrt{3}}-\sqrt{3}=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{3}=3-\sqrt{3}-\sqrt{3}=3-2\sqrt{3}\)
b) \(\sqrt{7+2\sqrt{6}}-\sqrt{3}=\sqrt{\left(1+\sqrt{6}\right)^2}-\sqrt{3}=1+\sqrt{6}-\sqrt{3}\)
- \(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)
- \(B=\left(\sqrt{28}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}=\left(2\sqrt{7}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}\)
\(=\left(3\sqrt{7}-4\right).\sqrt{7}+7\sqrt{7}=3\sqrt{7}+3\sqrt{7}=6\sqrt{7}\)
- \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
- \(D=0,2.\sqrt{10^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{3}-\sqrt{5}\right)=4\sqrt{3}-2\sqrt{5}\)
a: \(=2\cdot3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
b: \(=5\sqrt{10}+2\cdot5-5\sqrt{10}=10\)
c: \(=2\sqrt{7}\cdot\sqrt{7}-\sqrt{12}\cdot\sqrt{7}-\sqrt{7}\cdot\sqrt{7}+2\sqrt{21}=2\cdot7-7=7\)
d: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}=2\cdot11=22\)
a) ĐKXĐ : \(0\le a\ne1\)
\(\frac{\sqrt{a}-a}{\sqrt{a}-1}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
b) ĐKXĐ : \(b\ne0,a\ne-\sqrt{b}\)
\(\frac{a-\sqrt{b}}{\sqrt{b}}:\frac{\sqrt{b}}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\sqrt{b}}.\frac{a+\sqrt{b}}{\sqrt{b}}=\frac{a^2-b}{b}=\frac{a^2}{b}-1\)
c) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=\sqrt{5}\left(2-5-4+11\right)\)\(=4\sqrt{5}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
\(=7\left(2-2\sqrt{2}+1\right)+14\sqrt{2}=7\left(2-2\sqrt{2}+1+2\sqrt{2}\right)=7.3=21\)
e) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)