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\(3-x\times\frac{1}{2}=\frac{3}{4}\)
\(x\times\frac{1}{2}=3-\frac{3}{4}\)
\(x\times\frac{1}{2}=\frac{12}{4}-\frac{3}{4}\)
\(x\times\frac{1}{2}=\frac{9}{4}\)
\(x=\frac{9}{4}\div\frac{1}{2}\)
\(x=\frac{9}{4}\times\frac{2}{1}\)
\(x=\frac{18}{4}=\frac{9}{2}\)
Vậy x = 9/2
ik r mk làm tiếp cho
\(\left(a\right)37064-64\left(82+\frac{42966}{217}\right)\)
\(=37064-64\left(82+198\right)\)
\(=37064-64\cdot280\)
\(=30764-17920=19144\)
\(\left(b\right)320-\left(120,5+95,25+5,25\right)+\frac{84}{12}\cdot12,5\)
\(=320-\left(120,5+100,5\right)+7\cdot12,5\)
\(=320-221+87,5=11,5\)
\(\left(c\right)\left(4\frac{2}{5}+2\frac{3}{7}\right)-\left(2\frac{2}{5}-5\frac{4}{7}\right)\)
\(=4\frac{2}{5}-2\frac{2}{5}+2\frac{3}{7}+5\frac{4}{7}\)
\(=2+8=10\)
\(\left(d\right)\left(\frac{1998}{18}-\frac{1443}{13}\right)\left(16996-\frac{1110}{30}\cdot305\right)\)
\(=\left(111-111\right)\left(16996-\frac{1110}{30}\cdot305\right)\)
\(=0\left(16996-\frac{1110}{30}\cdot305\right)=0\)
\(\left(e\right)5\frac{3}{5}+1,75+6\frac{1}{8}+4\frac{1}{4}+3,875+3,4\)
\(=5\frac{3}{5}+1\frac{3}{4}+6\frac{1}{8}+4\frac{1}{4}+3\frac{7}{8}+3\frac{2}{5}\)
\(=5\frac{3}{5}+3\frac{2}{5}+1\frac{3}{4}+4\frac{1}{4}+6\frac{1}{8}+3\frac{7}{8}\)
\(=9+6+10=25\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
1.
a,\(\left(2013\times2014+2014\times2015+2015\times2016\right)\times\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2013\times2014+2014\times2015+2015\times2016\right)\times\left(1\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2013\times2014+2014\times2015+2015\times2016\right)\times0\)
\(=0\)
b, \(17,75+16,25+14,75+13,25+...+4,25+2,75+1,25\)
\(=\left(17,75+1,25\right)+\left(16,25+2,75\right)+...+9,75\)
\(=19\times7+9,75\)
\(=142,75\)
Hok Tốt!!!!
1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
\(\frac{22}{21}\): ( \(\frac{63}{4}\)\(-\frac{61}{4}\)) \(+\) \(\frac{25}{12}\): ( \(\frac{31}{4}\)\(-\)\(\frac{29}{4}\))
= \(\frac{22}{21}\): \(\frac{1}{2}\)+ \(\frac{25}{12}\): \(\frac{1}{2}\)
= ( \(\frac{22}{21}\)+ \(\frac{25}{12}\) ) : \(\frac{1}{2}\)
= \(\frac{263}{84}\) : \(\frac{1}{2}\)
=\(\frac{263}{42}\)
\(1\frac{1}{21}:\left(15,75-15\frac{1}{4}\right)+2\frac{1}{12}:\left(7\frac{3}{4}-7,25\right)\)
=\(\frac{22}{21}:\left(\frac{63}{4}-\frac{61}{4}\right)+\frac{25}{12}:\left(\frac{31}{4}-\frac{29}{4}\right)\)
=\(\frac{22}{21}:\frac{1}{2}+\frac{25}{12}:\frac{1}{2}=\frac{22x2}{21}+\frac{25x2}{12}\)
=\(\frac{44}{21}+\frac{25}{6}=\frac{88+175}{42}=\frac{263}{42}\)
b, 125 - 25:3 x 12 = 125 - \(\frac{25}{3}\)x 12 = 125 - 100 = 25
a, 50% + \(\frac{7}{12}\)- \(\frac{1}{2}\) = \(\frac{50}{100}+\frac{7}{12}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}+\frac{7}{12}=0+\frac{7}{12}=\frac{7}{12}\)