tính :
\(B=2\sqrt{\frac{0,01}{1,21}}+3\frac{2}{\sqrt{10^2+2^2+40}}-\frac{3}{4}\)
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6 tháng 12 2017
a) \(A=\dfrac{1}{\sqrt{25}}+\dfrac{\sqrt{49}}{\sqrt{36}}-\dfrac{2}{\sqrt{100}}.\)
\(=\dfrac{1}{5}+\dfrac{7}{6}-\dfrac{1}{5}.\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\dfrac{7}{6}.\)
\(=0+\dfrac{7}{6}=\dfrac{7}{6}.\)
Vậy \(A=\dfrac{7}{6}.\)
b) \(B=\sqrt{\dfrac{0,01}{1,21}}+3.\dfrac{2}{\sqrt{10^2}+2^2+40}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+3.\dfrac{2}{10+4+40}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+3.\dfrac{1}{37}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+\dfrac{1}{9}-\dfrac{3}{4}.\)
\(=\dfrac{36}{396}+\dfrac{44}{396}-\dfrac{297}{296}.\)
\(=-\dfrac{217}{396}.\)
Vậy \(B=-\dfrac{217}{396}.\)
28 tháng 6 2018
\(P=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(P=1+\sqrt{2}\)
bởi vì tách \(4=\sqrt{4}+\sqrt{4}\)
các bài khác tương tự
21 tháng 3 2020
\(B=\left(\frac{10\sqrt{1,21}}{7}+\frac{22\sqrt{0,25}}{3}\right):\left(\frac{5}{\sqrt{49}}+\frac{\sqrt{225}}{9}\right)\)
\(=\left(\frac{10.1,1}{7}+\frac{22.0,5}{3}\right):\left(\frac{3}{7}+\frac{15}{9}\right)\)
\(=\left(\frac{11}{7}+\frac{11}{3}\right):\left(\frac{27}{63}+\frac{105}{63}\right)\)
\(=\left(\frac{33}{21}+\frac{77}{21}\right):\frac{132}{63}\)
\(=\frac{110}{21}:\frac{132}{63}\)
\(=\frac{110}{21}.\frac{63}{132}\)
\(=\frac{5}{2}\)
Đúng ko nhỉ???
22 tháng 10 2021
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)
PN
19 tháng 8 2020
\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)
\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)
\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)
\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)}}+\frac{1}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)
Nếu \(3\frac{2}{\sqrt{10^2+2^2+40}}\)là hỗn số
=> B = \(2\sqrt{\frac{0,01}{1,21}}+3\frac{2}{\sqrt{10^2+2^2+40}}-\frac{3}{4}\)
B = \(2\sqrt{\frac{1}{121}}+3\frac{2}{144}-\frac{3}{4}\)
B = \(\frac{2}{11}+3\frac{1}{6}-\frac{3}{4}\)
B = \(\frac{2}{11}+\frac{19}{6}-\frac{3}{4}\)
B = \(\frac{343}{132}\)
tớ nói đơn giản thui:
\(\sqrt{a^2}=a\)