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B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
a) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\left(4+\sqrt{3}\right)+2\sqrt{3}\)
\(=3.2+4=6+4=10\)
b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)
c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)
d) \(=\frac{3\left(3-2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}+\frac{3\left(3+2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}=-6\)
e) Tự làm.
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
\(A=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{25-24}\right)\left(15+2\sqrt{6}\right)\)
\(A=10\left(15+2\sqrt{6}\right)\)
\(A=150+20\sqrt{6}\)
\(D=\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right).\frac{\sqrt{3}-1}{3-\sqrt{3}}=\left(\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{-1}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{-1}\right).\frac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\left(-5-2\sqrt{6}+5-2\sqrt{6}\right).\frac{1}{\sqrt{3}}=\frac{-4\sqrt{6}}{\sqrt{3}}=-4\sqrt{2}\)
\(E=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}=\frac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2}{\sqrt{2}}\)
\(=\sqrt{2}\)