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Lời giải:
Đặt biểu thức là $A$.
$21+4\sqrt{5}=20+1+2.\sqrt{20}.\sqrt{1}=(\sqrt{20}+1)^2$
$\Rightarrow \sqrt{21+4\sqrt{5}}=\sqrt{20}+1$
Do đó:
$A=\frac{11}{4-\sqrt{5}}+\frac{4}{3-\sqrt{5}}-\frac{19}{\sqrt{20}+1}$
$=\frac{11(4+\sqrt{5})}{(4-\sqrt{5})(4+\sqrt{5})}+\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{19(\sqrt{20}-1)}{(\sqrt{20}+1)(\sqrt{20}-1)}$
$=\frac{11(4+\sqrt{5})}{11}+\frac{4(3+\sqrt{5})}{4}-\frac{19(\sqrt{20}-1)}{19}$
$=4+\sqrt{5}+3+\sqrt{5}-(\sqrt{20}-1)$
$=8$
\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)
Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)
\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
a, \(\sqrt{2}A=\sqrt{10-2\sqrt{3.7}}+\sqrt{10+2\sqrt{3.7}}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{3}+\sqrt{7}=2\sqrt{7}\)
\(\Rightarrow A=\sqrt{14}\)
b, \(B=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{2}\)
c, \(C=\left(1-\sqrt{11}\right)\left(\sqrt{11}+1\right)=1-11=-10\)
d, \(D=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}-\frac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}\)
\(=-2-\sqrt{6}+2-\sqrt{6}=-2\sqrt{6}\)
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/ \(\frac{10}{\sqrt{\left(\sqrt{5}-2\right)^2}}-\frac{12}{\sqrt{\left(3+\sqrt{5}\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{3+\sqrt{5}}+\frac{20}{\sqrt{5}-1}\)
\(=\frac{10\left(\sqrt{5}+2\right)}{1}-\frac{12\left(3-\sqrt{5}\right)}{4}+\frac{20\left(\sqrt{5}+1\right)}{4}=16+18\sqrt{5}\)
\(\frac{10}{\sqrt{5}-2.\sqrt{5}.2+4}-\frac{12}{\sqrt{\sqrt{5}+2.\sqrt{5}.3+9}}+\frac{20}{\sqrt{5-2.\sqrt{5}.1+1}}=\frac{10}{\left(\sqrt{5}-2\right)^2}-\frac{12}{\sqrt{\left(\sqrt{5}+3\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{\sqrt{5}+3}+\frac{20}{\sqrt{5}-1}=\frac{10\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right).\left(\sqrt{5}+2\right)}-\frac{12.\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right).\sqrt{5}-3\left(\right)}+\frac{20.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\frac{10\sqrt{5}-20}{5-4}-\frac{12\sqrt{5}-36}{5-9}+\frac{20\sqrt{5}+20}{5-1}\\=\frac{40\sqrt{5}-80+12\sqrt{5}+36+20\sqrt{5}+20}{4}=\\ 18\sqrt{5}-6\)
\(\frac{11}{4-\sqrt{5}}+\frac{4}{3-\sqrt{5}}-\frac{19}{\sqrt{21+4\sqrt{5}}}=\frac{11\left(4+\sqrt{5}\right)}{16-5}+\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{19}{\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}+1}}\)
\(=4+\sqrt{5}+3+\sqrt{5}-\frac{19}{\sqrt{\left(2\sqrt{5}+1\right)^2}}=7+2\sqrt{5}-\frac{19}{2\sqrt{5}+1}\)
\(=7+2\sqrt{5}-\frac{19\left(2\sqrt{5}-1\right)}{20-1}=7+2\sqrt{5}-\left(2\sqrt{5}-1\right)=8\)