3+5+7+9+...+37
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d=43/37d=43/37 × 23/15−43/3723/15−43/37 × 8/15+31/378/15+31/37
=43/37=43/37 x 23/15+43/3723/15+43/37 x (−8/15)+31/37(−8/15)+31/37
=43/37(23/15−8/15)+31/37=43/37(23/15−8/15)+31/37
=43/37.1+31/37=43/37.1+31/37
=2=2
E=−5/24E=−5/24 × 9/11−5/249/11−5/24 × 2/11+7/122/11+7/12
=−5/24(9/11+2/11)+7/12=−5/24(9/11+2/11)+7/12
=−5/24.1+7/12=−5/24.1+7/12
=3/8=3/8
a)3/7–−11/14−7/3×6/7a)3/7–−11/14−7/3×6/7
=3/7−−11/14−2=3/7−−11/14−2
=3/7+11/14−2=3/7+11/14−2
=−11/14=−11/14
b)6/5:3/2+−2/3×9/14b)6/5:3/2+−2/3×9/14
=4/5−3/7=4/5−3/7
=13/35
d=43/37d=43/37 × 23/15−43/3723/15−43/37 × 8/15+31/378/15+31/37
=43/37=43/37 x 23/15+43/3723/15+43/37 x (−8/15)+31/37(−8/15)+31/37
=43/37(23/15−8/15)+31/37=43/37(23/15−8/15)+31/37
=43/37.1+31/37=43/37.1+31/37
=2=2
E=−5/24E=−5/24 × 9/11−5/249/11−5/24 × 2/11+7/122/11+7/12
=−5/24(9/11+2/11)+7/12=−5/24(9/11+2/11)+7/12
=−5/24.1+7/12=−5/24.1+7/12
=3/8=3/8
a)3/7–−11/14−7/3×6/7a)3/7–−11/14−7/3×6/7
=3/7−−11/14−2=3/7−−11/14−2
=3/7+11/14−2=3/7+11/14−2
=−11/14=−11/14
b)6/5:3/2+−2/3×9/14b)6/5:3/2+−2/3×9/14
=4/5−3/7=4/5−3/7
=13/35
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(a,\dfrac{47}{3}:3-\dfrac{37}{3}:3+\dfrac{5}{3}=\dfrac{1}{3}\times\left(\dfrac{47}{3}-\dfrac{37}{3}\right)+\dfrac{5}{3}=\dfrac{10}{9}+\dfrac{5}{3}=\dfrac{10}{9}+\dfrac{15}{9}=\dfrac{25}{9}\)
\(b,\dfrac{3}{5}:\dfrac{7}{9}\times\dfrac{7}{9}:2=\dfrac{27}{35}\times\dfrac{7}{18}=\dfrac{3}{10}\)
a) \(\frac{2}{9}+\frac{1}{5}+\frac{7}{9}+\frac{4}{5}=\left(\frac{2}{9}+\frac{7}{9}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)= 1+1=2 \)
b) \(\frac{19}{37}+\left(1+\frac{19}{37}\right)=\left(\frac{19}{37}-\frac{19}{37}\right)+1=0+1=1\)
c) \(\frac{9}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)+\frac{7}{8}=\frac{9}{8}+\frac{7}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)=2-2=0\)
d) \(\frac{3}{5}\times\frac{8}{27}\times\frac{5}{3}=\frac{3}{5}\times\frac{5}{3}\times\frac{8}{27}=1\times\frac{8}{27}=\frac{8}{27}\)
e) \(\frac{7}{19}\times\frac{1}{3}+\frac{7}{19}\times\frac{2}{3}=\frac{7}{19}\times\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}\times1=\frac{7}{19}\)
f) \(\frac{12}{5}\times4-4\times\frac{7}{5}=4\times\left(\frac{12}{5}-\frac{7}{5}\right)=4\times1=4\)
a) \(\frac{47}{3}:3-\frac{37}{3}:3+\frac{5}{3}=\left(\frac{47}{3}-\frac{37}{3}\right):3+\frac{5}{3}\)
\(=\frac{10}{6}+\frac{5}{3}=\frac{10}{3}\)
b) \(\frac{3}{5}:\frac{7}{9}.\frac{7}{9}+\frac{7}{5}:2=\frac{3}{5}+\frac{7}{10}=\frac{13}{10}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{37.39}\)
\(=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{37}-\frac{1}{39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(=\frac{1}{2}.\frac{4}{13}=\frac{2}{13}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(A=\frac{1}{2}.\frac{4}{13}\)
\(A=\frac{2}{13}\)
_Chúc bạn học tốt_
1) \(373737.73-37.737373\)
\(=37.10101.73-37.73.10101\)
\(=0\)
2) \(19.41414141-41.19191919\)
\(=19.41.1010101-41.19.1010101\)
\(=0\)
3) \(123456.456789456789-456789.123456123456\)
\(=123456.456789.1000001-456789.123456.1000001\)
\(=0\)
4) \(250.12.7\)
\(=250.4.3.7\)
\(=\left(250.4\right).\left(3.7\right)\)
\(=1000.21\)
\(=21000\)
5) \(24.125.9\)
\(=3.8.125.9\)
\(=\left(3.9\right).\left(8.125\right)\)
\(=27.1000\)
\(=27000\)
6) \(125.44.18\)
\(=125.4.11.2.9\)
\(=\left(125.4.2\right).\left(11.9\right)\)
\(=1000.99\)
\(=99000\)
7) \(8.24.6+12.4.52+2.22.24\)
\(=48.24+48.52+48.22\)
\(=48\left(24+52+22\right)\)
\(=48.98\)
\(=48\left(100-2\right)\)
\(=48.100-48.2\)
\(=4800-96\)
\(=4704\)
8) \(18.58.2-9.27.4+12.69.3\)
\(=36.58-36.27+36.69\)
\(=36\left(58-27+69\right)\)
\(=36.100=3600\)
9) \(30.85.3+18.72.5-45.37.2-9.20.10\)
\(=90.85+90.72-90.37-90.20\)
\(=90\left(85+72-37-20\right)\)
\(=90.100=9000\)
Khoảng cách : `2`
Số hạng :
`(37 - 3) : 2 + 1 = 18 (số - hạng)`
Tổng :
`(37 + 3) xx 18 : 2 = 360`
3 + 55 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37