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a) \(E=\sqrt{\left|12\sqrt{5}-29\right|}-\sqrt{12\sqrt{5}+29}\)
\(\Leftrightarrow E^2=\left|12\sqrt{5}-29\right|-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=29-12\sqrt{5}-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=-24\sqrt{5}\)
\(\Leftrightarrow E=-2\sqrt{6\sqrt{5}}\)
b) Đặt \(F=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)
\(\Leftrightarrow F^2=\left|40\sqrt{2}-57\right|-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=57-40\sqrt{2}-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=-80\sqrt{2}\)
\(\Leftrightarrow F=-4\sqrt{5\sqrt{2}}\)
ưu tiên phương pháp bình phương :
a) \(\left(4+\sqrt{15}\right)^2\left(\sqrt{10}-\sqrt{6}\right)^2\left(\sqrt{4-\sqrt{15}}\right)^2\)
\(=\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)^2\)
Tính ra kết quả nhớ căn đó
b) Phương pháp trục căn thức :
\(\frac{\sqrt{3+\sqrt{5}}\sqrt{3-\sqrt{5}}}{\sqrt{3-\sqrt{5}}}-\frac{\sqrt{3-\sqrt{5}}\sqrt{3+\sqrt{5}}}{\sqrt{3+\sqrt{5}}}-\sqrt{2}\)
Trên tử có hàng đẳng thức . bạn tự quy động là ra
a, \(\frac{\sqrt{3-\sqrt{5}}\times''3+\sqrt{5}''}{\sqrt{10}+\sqrt{2}}\)
\(=\frac{-9.976153125}{4.576491223}\)
b,\(\frac{''\sqrt{5}+2''^2-8\sqrt{5}}{2\sqrt{5}-4}\)
\(=\frac{0.05572809}{0.472135955}\)
P/s; Em không chắc đâu ạ. Mới lớp 5 lên 6 thôi
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
a) Ta có: \(\left(4+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\cdot\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{5}-2\sqrt{75}\)
b) Ta có: \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}-1\right)\cdot\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+3\right)\)
\(=\left(6-2\sqrt{5}\right)\cdot\left(\sqrt{5}+3\right)\)
\(=2\cdot\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)\)
\(=2\cdot\left(9-5\right)=2\cdot4=8\)
26, đặt bthuc là A suy ra A2=4+4+2\(\sqrt{16-\left(10+2\sqrt{5}\right)}\) suy ra A2=8+2(\(\sqrt{5}\) -1) suy ra A=\(\sqrt{6+2\sqrt{5}}\)=\(\sqrt{5}\)+1
40, tương tự
thanks p nhìu