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\(A=\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{15}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
\(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{4+\sqrt{7}}\sqrt{4-\sqrt{7}}-\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}-\left|4-\sqrt{7}\right|\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{16-7}-4+\sqrt{7}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=3-4+\sqrt{7}=-1+\sqrt{7}\)
\(\Leftrightarrow B=\frac{-1+\sqrt{7}}{\sqrt{4-\sqrt{7}}}\)
tíck mình nha bn thanks !!!!!!!!!!
\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)
\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)
\(=14+2\sqrt{49-40}=14+6=20\)
Khi đó:\(A=\sqrt{20}\)
Các câu còn lại bạn làm nốt nhé
a) \(A=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)
\(=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}\)
\(=\frac{12\sqrt{2}-2\sqrt{18}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)
\(A=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{25-24}\right)\left(15+2\sqrt{6}\right)\)
\(A=10\left(15+2\sqrt{6}\right)\)
\(A=150+20\sqrt{6}\)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6-2\sqrt{6}.\sqrt{7}+7}=\sqrt{\left(\sqrt{6}-\sqrt{7}\right)^2}=\left|\sqrt{6}-\sqrt{7}\right|=\sqrt{7}-\sqrt{6}\)
\(\sqrt{46+6\sqrt{5}}=\sqrt{45+6\sqrt{5}+1}=\sqrt{3^2.5+6\sqrt{5}+1}=\sqrt{3^2.5+2.3.\sqrt{5}+1^2}=\sqrt{\left(3.\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)
\(\sqrt{12-3\sqrt{15}}=\sqrt{3}\sqrt{4-\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{8-2\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{3-2\sqrt{15}+5}=\sqrt{\frac{3}{2}}.\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\frac{3}{2}}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{15}+5}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{3}\sqrt{5}+5}-\sqrt{3+2\sqrt{3}\sqrt{5}+5}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}=-2\sqrt{3}\)
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)=\sqrt{\frac{1}{2}}\left(1+\sqrt{5}-\sqrt{5}+1\right)=\sqrt{\frac{1}{2}}.2=\sqrt{\frac{4}{2}}=\sqrt{2}\)
Bạn xem lại đề. Biểu thức trong căn thứ 2 âm nên biểu thức B không tồn tại. Có phải số 8 bạn nên sửa thành 9?
uk mik sai đề