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kiểm tra bằng máy tính:
\(2\sqrt{2+\sqrt{50}\sqrt{18-\sqrt{128}}}>7\)
căn thức ko có nghĩa
\(\sqrt{7-2\sqrt{2+5\sqrt{2}+\sqrt{18-2\cdot4\cdot\sqrt{2}}}}\)=\(\sqrt{7-2\sqrt{2+5\sqrt{2}+4-\sqrt{2}}}\)
=\(\sqrt{7-2\sqrt{6+4\sqrt{2}}}=\sqrt{7-2\left(2+\sqrt{2}\right)}\) =\(\sqrt{3+2\sqrt{2}}\) =\(\sqrt{2}+1\)
\(\sqrt{18-\sqrt{128}}=\sqrt{18-8\sqrt{2}}=\sqrt{16-2.4.\sqrt{2}+2}=\sqrt{\left(4-\sqrt{2}\right)^2}=4-\sqrt{2}\)
=> \(\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}=\sqrt{2+5\sqrt{2}+4-\sqrt{2}}=\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{4+2.2\sqrt{2}+2}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
=> \(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{7-2\left(2+\sqrt{2}\right)}=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}.1+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
a) \(2\sqrt{50}-3\sqrt{32}-\sqrt{162}+5\sqrt{98}\)
=\(2.5\sqrt{2}-3.4\sqrt{2}-9\sqrt{2}+5.7\sqrt{2}\)
= \(10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}\)
= \(24\sqrt{2}\)
b) \(\sqrt{8+2\sqrt{7}}+\sqrt{11-4\sqrt{7}}\)
= \(\sqrt{7+2\sqrt{7}+1}+\sqrt{7-4\sqrt{7}+4}\)
= \(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-2\right)^2}\)
= \(\sqrt{7}+1+\sqrt{7}-2\)
= \(2\sqrt{7}-1\)
c) \(\dfrac{10}{\sqrt{5}}+\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{18}-3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
= \(2\sqrt{5}+6-2\sqrt{5}-3\)
= 3
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!
1,=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{12}+4}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
=\(\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{4+2\sqrt{3}}\)
=\(\sqrt{3}+1\)
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
= \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
= \(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{3}-2}\)
= \(\sqrt{4+2\sqrt{3}}\)
= \(\sqrt{3}+1\)