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\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(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) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
a,\(\sqrt{4\left(a-5\right)^2}=\sqrt{4}.\sqrt{\left(a-5\right)^2}=2.\left|a-5\right|=2\left(a-5\right)\left(a\ge5\right)\)
b,\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3=-1}\)
c,Mạn phép sửa đề ,nếu ko thì kết quả ko đẹp
\(\sqrt{8+2\sqrt{15}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{5}=\sqrt{5}+\sqrt{3}-\sqrt{5}=\sqrt{3}\)
d,\(\sqrt{\left(3-2\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=2\sqrt{3}-3-3-2\sqrt{3}=-6\)
e,\(\sqrt{24\left(b-3\right)}^2=\sqrt{24^2}.\sqrt{\left(b-3\right)^2}=24.\left(3-b\right)\left(b< 3\right)\)
\(\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(\sqrt{18}-\sqrt{20}+2\sqrt{2}\right)\)
\(=\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(3\sqrt{2}-2\sqrt{5}+2\sqrt{2}\right)\)
\(=\left(5\sqrt{2}-\sqrt{5}\right)\left(5\sqrt{2}-2\sqrt{5}\right)\)
\(=50-10\sqrt{10}-5\sqrt{10}+10\)
\(=60-15\sqrt{10}\)
\(\left(1+\sqrt{2}-\sqrt{5}\right)\left(1+\sqrt{2}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{2}\right)^2-5\)
\(=1+2\sqrt{2}+2-5\)
\(2\sqrt{2}-2\)
a) \(\sqrt{26+15\sqrt{3}}\)
\(=\frac{\sqrt{52+30\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.5+5^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{3}+5\right)^2}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}\)
b) \(\)\(\sqrt{2-\sqrt{3}}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}\)
c) \(\left(\sqrt{10}-\sqrt{2}\right).\left(\sqrt{3+5}\right)\)
\(=\sqrt{10}.\sqrt{8}-\sqrt{2}.\sqrt{8}\)
\(=\sqrt{80}-\sqrt{16}=4\sqrt{5}-4\)
d) \(\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}\)
\(=\left(\sqrt{6}-2\right)\left(\sqrt{5+\sqrt{24}}\right).\sqrt{5-\sqrt{24}}.\left(\sqrt{5+\sqrt{24}}\right)\)
\(=\left(\sqrt{6}-2\right)\left(\sqrt{5+\sqrt{24}}\right).1\)
\(=\left(\sqrt{6}-2\right).\left(\sqrt{5+\sqrt{24}}\right)\)
\(=\sqrt{2}.\left(\sqrt{3}-\sqrt{2}\right).\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{2}.\left(3-2\right)=\sqrt{2}\)
\(\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\\ =\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|\\ =5-\sqrt{24}-5-\sqrt{24}\\ =-2\sqrt{24}=-4\sqrt{6}\)
`\sqrt((5-\sqrt24)^2) - \sqrt((5+\sqrt24)^2)`
`=|5-\sqrt24|-|5+\sqrt24|`
`=5-\sqrt24-5-\sqrt24`
`=-2\sqrt24`
`=-4\sqrt6`