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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
mik làm bài này
linh tinh
bn ơi
cho mik
xin 1 L-I-K-E
b,
d,
\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{\sqrt{5}-2}-\frac{2}{2+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}\)
\(=2\sqrt{5}+4-2\sqrt{5}+4\)
\(=8\)
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=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
\(\frac{\sqrt{2-\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right).\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{4}:\left(\frac{2\sqrt{2+\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{2\sqrt{2+\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{4}:\left(\frac{\sqrt{4+2\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{4}:\left[\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4\sqrt{3}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\left[\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{4\sqrt{6}}-\frac{2.4}{4\sqrt{6}}+\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4\sqrt{6}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\frac{\sqrt{18}+\sqrt{6}-8+\sqrt{6}+\sqrt{2}}{4\sqrt{6}}\)
\(=\frac{\sqrt{3}-1}{4}.\frac{4\sqrt{6}}{\sqrt{2}\left(\sqrt{9}+2\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}+1\right)^2}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)^2}\)............
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{6\left(2+\sqrt{3}\right)}-4+\sqrt{2\left(2+\sqrt{3}\right)}}{2\sqrt{6}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}.\left(\frac{2\sqrt{6}}{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}\right)\)
\(=\frac{\sqrt{6\left(2+\sqrt{3}\right)}}{\left|\sqrt{3}+3\right|-4+\left|\sqrt{3}+1\right|}\)
\(=\frac{\left|\sqrt{3}+3\right|}{\sqrt{3}+3-4+\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+3}{2\sqrt{3}}\)
\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7-2\sqrt{10}}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}\right)^2+6\sqrt{2}+9}-\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{6}+\left(\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{10}+\left(\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+3-\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}+\sqrt{2}}\)
\(=\frac{3}{2\sqrt{2}+1}\)