tình GTBT \(P=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
Những câu hỏi liên quan
\(\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
Đặt: \(A=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}>0\)
<=> \(A.\sqrt{4+\sqrt{13}}=\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}\)
<=> \(A^2\left(4+\sqrt{13}\right)=4+\sqrt{3}+4-\sqrt{3}+2\sqrt{13}\)
<=> \(A^2\left(4+\sqrt{13}\right)=2\left(4+\sqrt{13}\right)\)
<=> \(A=\sqrt{2}\)
Vậy: \(\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
\(=\sqrt{2}+\sqrt{25-2.5.\sqrt{2}+2}\)
\(=\sqrt{2}+\left(5-\sqrt{2}\right)=5\)
Tính
T=\(\frac{\sqrt{4+\sqrt{3}+\sqrt{4-\sqrt{3}}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
Em tham khảo đề bài và bài làm tại link: Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath
Tính GTBT:
\(B=\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
=> B = \(\sqrt{2}\)
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Tính:
\(A=\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2014}}\)
\(B=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
Tính GTBT:
\(B=\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
\(B=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{8+2\sqrt{7}}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{8-2\sqrt{7}}}=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{\left(\sqrt{7}+1\right)^2}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{\left(\sqrt{7}-1\right)^2}}\)
\(=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{7+\sqrt{7}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{7-\sqrt{7}}=\frac{\sqrt{2}\left(4+\sqrt{7}\right)\left(7-\sqrt{7}\right)}{35}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)\left(7+\sqrt{7}\right)}{35}\)
\(=\frac{21\sqrt{2}+3\sqrt{14}}{35}+\frac{21\sqrt{2}-3\sqrt{14}}{35}=\frac{42\sqrt{2}}{35}=\frac{6\sqrt{2}}{5}\)
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Tínha.left(sqrt{10}-sqrt{2}right)left(3+sqrt{5}right)sqrt{27-9sqrt{5}}b.sqrt{frac{3sqrt{3}-4}{2sqrt{3}+1}}-sqrt{frac{4+sqrt{3}}{5-2sqrt{3}}}c.frac{3-4sqrt{3}}{sqrt{6}-sqrt{2}-sqrt{5}}d.left(sqrt{11}-sqrt{3}right)left(sqrt{13-sqrt{6}+2sqrt{30-sqrt{45}}}+sqrt{11}-sqrt{10-sqrt{6}}right)e.frac{sqrt{4+sqrt{5}}+sqrt{4-sqrt{5}}}{sqrt{4}+sqrt{11}}-frac{sqrt{20-4sqrt{23}}}{sqrt{5+sqrt{2}}-sqrt{5-sqrt{2}}}
Đọc tiếp
\(Tính\)
\(a.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\sqrt{27-9\sqrt{5}}\)
\(b.\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(c.\frac{3-4\sqrt{3}}{\sqrt{6}-\sqrt{2}-\sqrt{5}}\)
\(d.\left(\sqrt{11}-\sqrt{3}\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{45}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
\(e.\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4}+\sqrt{11}}-\frac{\sqrt{20-4\sqrt{23}}}{\sqrt{5+\sqrt{2}}-\sqrt{5-\sqrt{2}}}\)
rút gọn biểu thức :
A= \(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\).
B= \(\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\).
C= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\).
Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
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Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)
=1
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Tìm giá trị của biểu thức N=\(\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)
Giải phương trình x2 - x - 4 = 2\(\sqrt{x-1}\)(1-x)
16 tháng 10 2021 lúc 21:34
giác trị của biểu thức N=A= \(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)