\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{8-\sqrt{128}}}}\)
Những câu hỏi liên quan
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2^2}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\left|4-\sqrt{2}\right|}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left|\sqrt{3}-1\right|}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{3}-2}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\sqrt{3^2}-1^2\\ =3-1\\ =2\)
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\(\sqrt{6+2\sqrt{2}\sqrt{3+\sqrt{\sqrt{2+\sqrt{12+\sqrt{18-\sqrt{128}}}}}}}\)
\(\sqrt{6+2\sqrt{2}\sqrt{3+\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18-\sqrt{18-\sqrt{128}}}}}}}\)
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Rút gọn : Sử dụng công thức \(\sqrt{A^2}=\left|A\right|\)
a) \(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}\)
b) \(\sqrt{8}.\sqrt{3-\sqrt{5}}\)
c) \(\sqrt{15-6\sqrt{6}}-\sqrt{33-12\sqrt{6}}\)
d) \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
B=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3}+1}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{2-\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{4-2\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(B=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(B=\sqrt{4+2\sqrt{3}}\)
\(B=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(B=\sqrt{3}+1\)
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\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18}-\sqrt{128}}}}\)
B= \(\sqrt{6+2\sqrt{2.}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
rút gọn
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Ta có \(\sqrt{18-\sqrt{128}}\)
= \(\sqrt{18-8\sqrt{2}}\)
= \(\sqrt{16-2×4×\sqrt{2}+2}\)
= \(4-\sqrt{2}\)
Từ đó cái ban đầu
= \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
= \(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{3}-2}\)
= \(\sqrt{4+2\sqrt{3}}\)
= \(\sqrt{3}+1\)
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Xem thêm câu trả lời
Rút gọn : Sử dụng công thức \(\sqrt{A^2}=\left|A\right|\)
a) \(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}\)
b) \(\sqrt{8}.\sqrt{3-\sqrt{5}}\)
c) \(\sqrt{15-6\sqrt{6}}-\sqrt{33-12\sqrt{6}}\)
d) \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
a) \(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2}=\frac{\sqrt{3-2\sqrt{3}+1}}{2}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}\)
\(=\frac{\left|\sqrt{3}-1\right|}{2}=\frac{\sqrt{3}-1}{2}\)
b) \(\sqrt{8}\cdot\sqrt{3-\sqrt{5}}=\sqrt{4}\cdot\sqrt{6-2\sqrt{5}}=2\sqrt{5-2\sqrt{5}+1}=2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\cdot\left|\sqrt{5}-1\right|=2\left(\sqrt{5}-1\right)=2\sqrt{5}-2\)