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Đặt t\(^2\) = \(\sqrt{\text{2}}\)
=> \(\sqrt{\text{t^2 + 2t - 1}}\)+ \(\sqrt{\text{t^2 - 2t - 1}}\)
=> \(\sqrt{\text{t^2 + 2t + 1 - 2}}\)+ \(\sqrt{\text{t^2 - 2t + 1 - 2}}\)
=> \(\sqrt{\text{(t+ 1)^2 - 2}}\)+ \(\sqrt{\text{(t - 1)^2 - 2}}\)
Bạn làm mình k hiểu lắm, cũng chẳng biết đúng k, nhưng bài của mình làm ra rồi. nên mình k cho bạn vậy
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
b)\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
\(\Leftrightarrow\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)
\(\Leftrightarrow\sqrt{m-1}+1+\sqrt{m-1}-1\Leftrightarrow2\sqrt{m-1}\)
Câu 1 phá từng lớp ra :VD\(9+4\sqrt{2}\) =\((\sqrt{2}+2)^2\)
Câu 2:m+2\(\sqrt{m-1}\) =m-1+1+2\(\sqrt{m-1}\) =\((\sqrt{m-1} -1)^2 \)
- \(\frac{2+\sqrt{2}}{1+\sqrt{2}}=\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{2}\)
- \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=-\sqrt{5}\)
- \(\frac{2\sqrt{3}-\sqrt{6}}{1-\sqrt{3}}=\frac{-\sqrt{6}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=-\sqrt{6}\)
- \(\frac{a-\sqrt{a}}{1-\sqrt{a}}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
- \(\frac{p-2\sqrt{p}}{\sqrt{p}-2}=\frac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}.\)
\(=\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\left(\sqrt{3}-1\right)}+\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\left(\sqrt{3}+1\right)}+\frac{5}{\sqrt{6}}\)
\(=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{3-1}+\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{3+1}+\frac{5}{\sqrt{6}}\)
\(=\frac{\left(\sqrt{3}+1\right)}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{8}}+\frac{5}{\sqrt{6}}\)
\(=...\)
\(a,\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
\(=\frac{2.\left(\sqrt{6}+2+\sqrt{6}-2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}+\frac{5\sqrt{6}}{6}\)
\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}\)
\(=\frac{17\sqrt{6}}{6}\)
\(b,\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}\)
\(=\frac{2\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\)
\(=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)
\(\sqrt{\sqrt{2}+2\sqrt{2-1}}+\)\(\sqrt{\sqrt{2}-2\sqrt{2-1}}\)
\(=\sqrt{\sqrt{2}+2\sqrt{1}}+\)\(\sqrt{\sqrt{2}-2.\sqrt{1}}\)
\(=\sqrt{\sqrt{2}+2.1}+\)\(\sqrt{\sqrt{2}-2.1}\)
\(=\sqrt{\sqrt{2}+2}+\)\(\sqrt{\sqrt{2}-2}\)
\(=\sqrt[4]{2}\sqrt{2}+\sqrt[4]{2}\left(-\sqrt{2}\right)\)
\(=\sqrt[4]{2}\left(\sqrt{2}+-\sqrt{2}\right)\)
\(=\sqrt[4]{2}.0\)
\(=0\)
Mk ko chắc đúng nên sai đừng chửi nhé
Dương lớp 6 chưa học thì đừng có làm
Phan Hoàng Quốc Khánh đề có sai không bạn ? \(\sqrt{\sqrt{2}-2\sqrt{2-1}}=\sqrt{\sqrt{2}-2}\)
mà \(\sqrt{2}< 2\)nên \(\sqrt{\sqrt{2}-2}\)không tồn tại
xem lại đề đi bạn :)