\(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+..+\frac{1}{\sqrt{9997}+\sqrt{9999}}>24\)
Chứng minh rằng: \(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{9997}+\sqrt{9999}}>24\)
chứng minh rằng A=\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{9997}+\sqrt{9999}}\)>24
Rút gọn các biểu thức sau:
1) \(\frac{1}{\sqrt{7-\sqrt{24}+1}}-\frac{1}{\sqrt{7+\sqrt{24}}}\)
2) \(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}\)
3) \(\sqrt{\frac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+\sqrt{6}}}\)
4) \(\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\)
Tính:
\(a)\frac{1}{\sqrt{7-\sqrt{24}}+1}+\frac{1}{\sqrt{7+\sqrt{24}}-1}\\ b)\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2014}+\sqrt{2015}}\)
\(a=\frac{1}{\sqrt{7-2\sqrt{6}}+1}+\frac{1}{\sqrt{7+2\sqrt{6}}-1}=\frac{1}{\sqrt{\left(\sqrt{6}-1\right)^2}+1}+\frac{1}{\sqrt{\left(\sqrt{6}+1\right)^2}-1}\)
\(=\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}\)
Coi lại đề câu b, quy luật ở số hạng cuối cùng sai (nhìn 2 số hạng đầu 2 số dưới căn hơn kém nhau 4 đơn vị, số cuối lại chỉ hơn kém nhau 1 đơn vị)
Tính
\(\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(\frac{1}{\sqrt{7-\sqrt{24}+1}}-\frac{1}{\sqrt{7+\sqrt{24}-1}}:\left(\sqrt{3}-\sqrt{2}\right)\)
Tinh gia tri cua bieu thuc : a) A = \(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
b) B = \(\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)
Bạn trục căn thức ở mẫu rồi trừ đi là xong nhé,vì khi trục căn thức thì ở A mẫu chung là 1,ở B mẫu chung là 2.
A=(√3-√2)/(3-2)+(√4-√3)/(4-3)+......
=√3-√2+√4-√3+......+√25-√24
=√25-√2=5-√2.Câu b tương tự
thực hiện phép tính:
1, \(\sqrt{5+2\sqrt{24}}-\sqrt{2}\)
2, \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}\) 3, \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\)
4, \(\frac{1}{1-\sqrt{2}}-\frac{1}{1+\sqrt{2}}\) 5, \(\frac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\frac{3-\sqrt{3}}{\sqrt{3}}-\frac{4}{1-\sqrt{7}}\)
a/ \(\sqrt{5+\sqrt{24}}-\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{2}=\left|\sqrt{3}+\sqrt{2}\right|-\sqrt{2}=\sqrt{3}+\sqrt{2}-\sqrt{2}=\sqrt{3}\)
b/ \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}=\sqrt{3}\)
c/ \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
d/ \(\frac{1}{1-\sqrt{2}}-\frac{1}{1+\sqrt{2}}=\frac{1+\sqrt{2}-1+\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\frac{2\sqrt{2}}{1-2}=-2\sqrt{2}\)
Bài 1. thực hiện phép tính
a) \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\) b) \(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
Bài 2. Tính:a) \(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
b) \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
c)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}0-5\sqrt{2}}\)
\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)
Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)
\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)
13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)
17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)
18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)
20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)
21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)
22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)
24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)
26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)
28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)
29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)