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\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)
Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)
\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\frac{1}{4}=\sqrt{3}-\frac{3}{4}\)
\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)
\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)
\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)
\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)
\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)}}+\frac{1}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)
Mk năm nay mới lên lớp 9 thôi nhưng cũng biết chút!Mk giải ho bạn câu 1 còn lại bạn tự giải nhé!
1,\(\frac{1}{1+\sqrt{5}}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{1}{\sqrt{5}+1}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)+\(\frac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
=\(\frac{\sqrt{5}-1}{5-1}\)+\(\frac{\sqrt{5}+1}{5-1}\)
=\(\frac{\sqrt{5}-1}{4}\)+\(\frac{\sqrt{5}+1}{4}\)
=\(\frac{\sqrt{5}-1+\sqrt{5}+1}{4}\)
=\(\frac{2\sqrt{5}}{4}\)
=\(\frac{\sqrt{5}}{2}\)
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{9}+\sqrt{10}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{10}-\sqrt{9}\)
\(=\sqrt{10}-1\)
\(B=\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{7}}+\frac{2}{\sqrt{7}+\sqrt{9}}\)
\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}\)
\(=\sqrt{9}-1\)