RÚT GỌN
\(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2Cos2048x}}}}}}}\)
Rút gọn : \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\sqrt{2}+2}\)
\(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\left(\sqrt{2}+1\right)}\)
\(=\sqrt{\sqrt{2}-1}-\left(\sqrt{2}-1\right)\sqrt{\sqrt{2}+1}\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}\left(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\right)\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}.1\)
\(=0\)
Rút gọn biểu thức
\(A=\sqrt{30+12\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)
\(B=\sqrt{2+\sqrt{2}}.\sqrt{2+\sqrt{2+2}}.\sqrt{2-\sqrt{2+2}}\)
\(A=\sqrt{\left(9\sqrt{2}+2\sqrt{3}\right)^2}-\sqrt{\left(9\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\left|9\sqrt{2}+2\sqrt{3}\right|-\left|9\sqrt{2}-\sqrt{3}\right|\)
\(=9\sqrt{2}+2\sqrt{3}-9\sqrt{2}+\sqrt{3}=3\sqrt{3}\)
Kiểm tra lại đề bài câu B, chỗ \(\sqrt{2+\sqrt{2+2}}\)
Nếu câu B sửa đề thành:
\(B=\sqrt{2+\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{2+\sqrt{2}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\)
\(=\sqrt{2+\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)
\(=\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{4-2}=\sqrt{2}\)
Rút gọn
\(\sqrt{2+\sqrt{3}}+\sqrt{2+\sqrt{2+\sqrt{3}}}+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
chịu,,, chắc toàn dấu cộng chứ tự nhiên có dấu trừ sao làm
Rút gọn :
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}+\dfrac{1}{\sqrt{2}+1}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2}+1-\sqrt{2}+1=2\)
Rút gọn:
\(A=\sqrt{3+2\sqrt{2}}-\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)
A = \(\sqrt{3+2\sqrt{2}}-\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{3+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}+3}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2+2\sqrt{3}\left(\sqrt{2}+1\right)+3}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}+1+\sqrt{3}\right)^2}\)
= \(\left|\sqrt{2}+1\right|-\left|\sqrt{2}+\sqrt{3}+1\right|\)
= \(\sqrt{2}+1-\sqrt{2}-\sqrt{3}-1\)
= \(-\sqrt{3}\)
5.Rút gọn:
B= \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}\)
6.Rút gọn:
A= \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Rút gọn biểu thức :
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
Rút gọn B=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\)\(.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
rút gọn
\(\sqrt(\sqrt(2)+2\sqrt(\sqrt(2)-1))+\sqrt(\sqrt(2)-2\sqrt(\sqrt(2)-1))\)