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1) Ta có: \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=1+\sqrt{2}\)
2) Ta có: \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
\(=\sqrt{108}-\sqrt{36\cdot\frac{4}{3}}+\sqrt{75\cdot\frac{9}{25}}\)
\(=\sqrt{108}-\sqrt{48}+\sqrt{27}\)
\(=\sqrt{3}\left(6-4+3\right)\)
\(=5\sqrt{3}\)
3) Sửa đề: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
Ta có: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
\(=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}-10\sqrt{4}\cdot\sqrt{3}+16\cdot\sqrt{4}\cdot\sqrt{3}\)
\(=\sqrt{2}\cdot\sqrt{12}-10\sqrt{12}+16\sqrt{12}\)
\(=\sqrt{12}\left(\sqrt{2}-10+16\right)\)
\(=2\sqrt{3}\left(\sqrt{2}-6\right)\)
\(=2\sqrt{6}-12\sqrt{3}\)
4) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{12}}{6}-\frac{2\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6\left(2-\sqrt{3}\right)+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{12-6\sqrt{3}+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{3}}\)
5) Ta có: \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\frac{\sqrt{3}\left(2+5+3\right)}{\sqrt{15}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)
6) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)
\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)
\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)
h)
\(H=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=2\sqrt{3}\)
i)
\(I=\frac{2+\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3.1}}}+\frac{2-\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3.1}}}=\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6}{6}=1\)
ê)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+1+2(\sqrt{2}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+1+2(\sqrt{2}+\sqrt{5})}=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}=\sqrt{2}+\sqrt{5}+1\)
g)
\(13+\sqrt{48}=13+2\sqrt{12}=12+1+2\sqrt{12.1}=(\sqrt{12}+1)^2\)
\(\Rightarrow \sqrt{13+\sqrt{48}}=\sqrt{12}+1\)
\(\Rightarrow \sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{4+\sqrt{12}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
\(\Rightarrow 2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}=2\sqrt{2-\sqrt{3}}=\sqrt{2}.\sqrt{4-2\sqrt{3}}=\sqrt{2}.\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{2}(\sqrt{3}-1)=\sqrt{6}-\sqrt{2}\)
\(\Rightarrow G=1\)
1/\(\sqrt{8-2\sqrt{15}}-\sqrt{21-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
Bạn tự làm tiếp
2/ \(\frac{4}{\sqrt{7-4\sqrt{3}}}-\frac{4}{7-4\sqrt{3}}=\frac{4}{\sqrt{\left(2-\sqrt{3}\right)^2}}-\frac{4}{\left(2-\sqrt{3}\right)^2}=\frac{4}{2-\sqrt{3}}-\frac{4}{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{8-4\sqrt{3}-4}{\left(2-\sqrt{3}\right)^2}=\frac{4-4\sqrt{3}}{\left(2-\sqrt{3}\right)^2}\) đến đây ko rút gọn được nữa, nghi bạn chép sai đề.
Tử số của phân số thứ hai là 4 hay 1 vậy?
3/ \(\frac{\sqrt{8+2\sqrt{15}}-\sqrt{4-2\sqrt{3}}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{3+\sqrt{5}}{2}\)
4/ \(\frac{10}{\sqrt{\left(\sqrt{5}-2\right)^2}}-\frac{12}{\sqrt{\left(3+\sqrt{5}\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{3+\sqrt{5}}+\frac{20}{\sqrt{5}-1}\)
\(=\frac{10\left(\sqrt{5}+2\right)}{1}-\frac{12\left(3-\sqrt{5}\right)}{4}+\frac{20\left(\sqrt{5}+1\right)}{4}=16+18\sqrt{5}\)
\(\frac{10}{\sqrt{5}-2.\sqrt{5}.2+4}-\frac{12}{\sqrt{\sqrt{5}+2.\sqrt{5}.3+9}}+\frac{20}{\sqrt{5-2.\sqrt{5}.1+1}}=\frac{10}{\left(\sqrt{5}-2\right)^2}-\frac{12}{\sqrt{\left(\sqrt{5}+3\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{\sqrt{5}+3}+\frac{20}{\sqrt{5}-1}=\frac{10\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right).\left(\sqrt{5}+2\right)}-\frac{12.\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right).\sqrt{5}-3\left(\right)}+\frac{20.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\frac{10\sqrt{5}-20}{5-4}-\frac{12\sqrt{5}-36}{5-9}+\frac{20\sqrt{5}+20}{5-1}\\=\frac{40\sqrt{5}-80+12\sqrt{5}+36+20\sqrt{5}+20}{4}=\\ 18\sqrt{5}-6\)
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