rút gọn:
\(\frac{9}{2\sqrt{7}+1}+\sqrt{\frac{2}{4-\sqrt{7}}}\)
Rút gọn : \(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
\(\frac{1}{\sqrt{n}-\sqrt{n+1}}=\frac{\sqrt{n}+\sqrt{n+1}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n+1}\right)}=-\sqrt{n}-\sqrt{n+1}\)
Rút gọn \(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}-\sqrt{9}}\)
Phân tích mỗi hạng tử theo kiểu như dưới đây
\(\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}\right)^2-\left(\sqrt{2}\right)^2}\)
\(\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}\)
Khi đó mọi mẫu đều bằng -1
Bạn tiếp tục làm và kết quả nhận được là \(1-\sqrt{9}\)
rút gọn
\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}\)
\(-\frac{1}{\sqrt{8}-\sqrt{9}}\)
\(\frac{1}{\sqrt{9}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{4}}\)rút gọn
rút gọn phân số sau:
A = \(\frac{1-\sqrt[1]{49}+\sqrt[1]{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\sqrt{\frac{64}{2}}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
Rút gọn biểu thức
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(=\frac{1}{4}\)
Rút gọn
1,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
2,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
3,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
4,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
5,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
1) Rút gọn biểu thức:
a) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
b) \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
c) \(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\)
d) \(\frac{4}{\sqrt{5}-\sqrt{2}}+\frac{3}{\sqrt{5}-2}-\frac{2}{\sqrt{3}-2}+\frac{\sqrt{3}-1}{6}\)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
rút gọn \(\frac{-3}{5+2\sqrt{7}}-\frac{1}{3+\sqrt{7}}+\frac{6}{\sqrt{7}-2}+\frac{5-\sqrt{7}}{2}\)
Đề : Rút gọn biểu thức
B =\(\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}\)
C =\(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
\(B=\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}-\sqrt{5}-\sqrt{7}}{5-7}=\frac{-2\sqrt{7}}{-2}=\sqrt{7}\)
\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}=\sqrt{\left(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)^2}\)
\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}+2\sqrt{\frac{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}}+\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
\(C=\sqrt{\frac{\left(4+\sqrt{7}\right)^2}{16-7}+\frac{\left(4-\sqrt{7}\right)^2}{16-7}+2}\)
\(C=\sqrt{\frac{\left(4+\sqrt{7}+4-\sqrt{7}\right)^2-2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{16-7}+2}\)
\(C=\sqrt{\frac{16^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{238}{9}+2}=\sqrt{\frac{256}{9}}=\frac{16}{3}\)
Chúc bạn học tốt ~
đoạn cuối sửa lại nhé -,- tính ngu
\(C=\sqrt{\frac{8^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{46}{9}+2}=\sqrt{\frac{64}{9}}=\frac{8}{3}\)
Chúc bạn học tốt ~