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\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)
\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)
\(=\left(\left(-\sqrt{7}\right)+\left(-\sqrt{5}\right)\right)\cdot\frac{\sqrt{7}-\sqrt{7}}{1}\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\cdot\frac{\sqrt{7}-\sqrt{5}}{1}\)
\(=\frac{-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{1}\)
\(=\frac{-\left(7-5\right)}{1}=-2\)
Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)
\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\frac{1}{4}=\sqrt{3}-\frac{3}{4}\)
\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)
\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)
\(2A=\frac{2}{1+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}\)
\(2A>\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)
Nhân liên hợp tử - mẫu vế phải:
\(\Rightarrow2A>\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)
\(2A>\frac{1}{2}\left(\sqrt{101}-1\right)>\frac{1}{2}\left(\sqrt{100}-1\right)=\frac{9}{2}\)
\(\Rightarrow A>\frac{9}{4}\)
\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}+1}}\)
\(=\frac{1}{\sqrt{6-2\sqrt{6}+1}+1}-\frac{1}{\sqrt{6+2\sqrt{6}+1}+1}\)
\(=\frac{1}{\sqrt{\left(\sqrt{6}-1\right)^2}+1}-\frac{1}{\sqrt{\left(\sqrt{6}+1\right)^2}+1}\)
\(=\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}+2}\)
\(=\frac{\sqrt{6}+2}{\sqrt{6}.\left(\sqrt{6}+2\right)}-\frac{\sqrt{6}}{\sqrt{6}.\left(\sqrt{6}+2\right)}\)
\(=\frac{2}{6+2\sqrt{6}}=\frac{12-4\sqrt{6}}{12}=\frac{3-\sqrt{6}}{3}\)
bài này dễ mà bạn cần mk giải chi tiết ko
kết quảA =\(\frac{\sqrt{99}-\sqrt{3}}{2}\)
Có: \(\frac{1}{\sqrt{n}+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}-\sqrt{n}\right)\left(\sqrt{n+2}+\sqrt{n}\right)}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
\(\Rightarrow A=\frac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{99}-\sqrt{97}}{2}\)
\(A=\frac{\sqrt{99}-\sqrt{3}}{2}\)
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}.\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}\)\(+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\frac{4+2\sqrt{3}}{4}}{1+\sqrt{\frac{4+\sqrt{3}}{4}}}\)\(+\frac{\frac{4-2\sqrt{3}}{4}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}\)
\(=\frac{\frac{3+2\sqrt{3}+1}{4}}{1+\sqrt{\frac{3+2\sqrt{3}+1}{4}}}\)\(+\frac{\frac{3-2\sqrt{3}+1}{4}}{1-\sqrt{\frac{3-2\sqrt{3}+1}{4}}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{2+\sqrt{3}}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{2-\sqrt{3}}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{\left(\sqrt{3}+1\right)^2}{4}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{\left(\sqrt{3}-1\right)^2}{4}}\)
\(=1+1=2\)
\(A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(A=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(1-\frac{\sqrt{3}}{2}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(A=\frac{\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\)
\(A=\frac{3+\sqrt{3}+3-\sqrt{3}}{6}\)
\(A=\frac{6}{6}=1\)
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.
giai ra giup mik di