E= \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{2+\sqrt{2+\sqrt{18+\sqrt{128}}}}}}\)là một số nguyên.
LÀM GIÚP MÌNH GẤP
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2^2}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\left|4-\sqrt{2}\right|}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left|\sqrt{3}-1\right|}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{3}-2}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\sqrt{3^2}-1^2\\ =3-1\\ =2\)
\(\sqrt{6+2\sqrt{2}\sqrt{3+\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18-\sqrt{18-\sqrt{128}}}}}}}\)
Chứng minh biểu thức sau nhận giá trị nguyên:
\(D=\left(\sqrt{3}-1\right)\sqrt{6+2.\sqrt{2.}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18}-\sqrt{128}}}}\)
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
giúp với
Ta có: \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
= \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2\cdot\sqrt{2}\cdot\sqrt{16}+2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(\sqrt{16}-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16}-\sqrt{2}}}\)
=\(\sqrt{6-2\sqrt{4+\sqrt{12}}}\)
=\(\sqrt{6-2\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}\)
=\(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}\cdot1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}\)=\(\sqrt{3}-1\)
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(16-\sqrt{2}\right)^2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2+\sqrt{12}+16-\sqrt{2}}}}\)
\(=\sqrt{6-2\sqrt{16+\sqrt{12}}}\) \(=\sqrt{6-2\sqrt{16+2\sqrt{3}}}\)
Thật xin lỗi! Phân tích đến đây là mk tịt r! Bn đok qa có khi lại nghĩ ra đấy!\(\sqrt{6+2\sqrt{2}\sqrt{3+\sqrt{\sqrt{2+\sqrt{12+\sqrt{18-\sqrt{128}}}}}}}\)
1) Rút gọn
h)\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)\)
e)\(\frac{\sqrt{28}-2\sqrt{12}-2\sqrt{18}}{3\sqrt{7}-2\sqrt{27}-\sqrt{102}}\)
f)\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
mọi ng giúp mình vs đang cần gấp tks !!!đc câu nào giúp câu đấy cx đk
\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)
\(=11.2.13.\sqrt{9}-1=286.3-1=857\)
\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)
\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)
\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)
B=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3}+1}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{2-\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{4-2\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(B=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(B=\sqrt{4+2\sqrt{3}}\)
\(B=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(B=\sqrt{3}+1\)
\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18}-\sqrt{128}}}}\)
B= \(\sqrt{6+2\sqrt{2.}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)