\(128\sqrt{e980}\) = ????
__ #mắm cấm trả lời nghe chưa...=.=
Những câu hỏi liên quan
\( {128 \sqrt{e980} \over }\) là gì?
(đố mẹo nha, có thể tìm kiếm trên mạng)
Xem thêm câu trả lời
\(128\sqrt{e980}\ \)
kết quả là bao nhiêu...toán khó à nha...ai giải dc cho 5 tích....(kết quả sẽ ra = chữ cái chứ ko ra số..^^)
IZ6\(\sqrt{e980}\)
Hi các bn , lm wen nha
123 x 2 = ?
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2^2}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\left|4-\sqrt{2}\right|}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left|\sqrt{3}-1\right|}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{3}-2}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\sqrt{3^2}-1^2\\ =3-1\\ =2\)
Đúng 3
Bình luận (0)
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{8-\sqrt{128}}}}\)
\(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
kiểm tra bằng máy tính:
\(2\sqrt{2+\sqrt{50}\sqrt{18-\sqrt{128}}}>7\)
căn thức ko có nghĩa
Đúng 0
Bình luận (0)
\(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
\(\sqrt{7-2\sqrt{2+5\sqrt{2}+\sqrt{18-2\cdot4\cdot\sqrt{2}}}}\)=\(\sqrt{7-2\sqrt{2+5\sqrt{2}+4-\sqrt{2}}}\)
=\(\sqrt{7-2\sqrt{6+4\sqrt{2}}}=\sqrt{7-2\left(2+\sqrt{2}\right)}\) =\(\sqrt{3+2\sqrt{2}}\) =\(\sqrt{2}+1\)
Đúng 0
Bình luận (0)
\(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
\(\sqrt{18-\sqrt{128}}=\sqrt{18-8\sqrt{2}}=\sqrt{16-2.4.\sqrt{2}+2}=\sqrt{\left(4-\sqrt{2}\right)^2}=4-\sqrt{2}\)
=> \(\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}=\sqrt{2+5\sqrt{2}+4-\sqrt{2}}=\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{4+2.2\sqrt{2}+2}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
=> \(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{7-2\left(2+\sqrt{2}\right)}=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}.1+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
Đúng 0
Bình luận (0)
\(\sqrt{6+2\sqrt{2}\sqrt{3+\sqrt{\sqrt{2+\sqrt{12+\sqrt{18-\sqrt{128}}}}}}}\)
