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12 tháng 1 2022
Bài 2
a: =>x=-23-7=-30
b: =>75:x=15
hay x=5
c: =>2(x+4)=80
=>x+4=40
hay x=36
d: \(\Leftrightarrow2^x\cdot11=88\)
hay x=3
Bài 3:
Có thể chia được nhiều nhất 12 tổ vì UCLN(180;132)=12
Khi đó, mỗi tổ có 15 nam và 11 nữ
11 tháng 12 2021
\(a,=35-2.1+3.7^3=33+3.343=33+1029=1062\\ b,=5.64+6-162+7=320-149=171\)
NH
22 tháng 8 2020
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)
22 tháng 8 2020
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)
NH
1
1 tháng 7 2015
\(\Leftrightarrow2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)+4x=7.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)\(\Leftrightarrow2.\left(1-\frac{1}{49}\right)+4x=7.\left(1-\frac{1}{99}\right)\)
\(\Leftrightarrow2.\frac{48}{49}+4x=7.\frac{98}{99}\)
\(\Leftrightarrow\frac{96}{49}+4x=\frac{686}{99}\)
\(\Leftrightarrow4x=\frac{686}{99}-\frac{96}{49}\)
\(\Leftrightarrow4x=4,970109256\)
\(\Leftrightarrow x=4,970109256:4\)
\(\Leftrightarrow x=1,242527314\)
AH
Akai Haruma
Giáo viên
15 tháng 1
a/
$\frac{7}{4}+\frac{9}{16}=\frac{28}{16}+\frac{9}{16}=\frac{28+9}{16}=\frac{37}{16}$
b/
$\frac{-70}{27}+\frac{35}{9}=\frac{-70}{27}+\frac{105}{27}=\frac{35}{27}$
c/
$\frac{-4}{-5}+\frac{-2}{3}=\frac{4}{5}-\frac{2}{3}=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}$
d/ Đề thiếu. Bạn xem lại.
i/
$\frac{-7}{2}+(\frac{-5}{-4}-\frac{4}{-9}$
$=\frac{-7}{2}+\frac{5}{4}+\frac{4}{9}$
$=\frac{-14}{4}+\frac{5}{4}+\frac{4}{9}$
$=\frac{-9}{4}+\frac{4}{9}=\frac{-81}{36}+\frac{16}{36}=\frac{-65}{36}$
AH
Akai Haruma
Giáo viên
15 tháng 1
k/
$\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$
$=\frac{3}{12}+\frac{4}{12}-\frac{6}{12}=\frac{1}{12}$
m/
$\frac{-1}{9}-\frac{1}{7}-\frac{3}{4}$
$=-(\frac{1}{9}+\frac{1}{7}+\frac{3}{4})$
$=-(\frac{28}{252}+\frac{36}{252}+\frac{189}{252})$
$=\frac{-253}{252}$
n/
$\frac{11}{12}-(\frac{-3}{18}+\frac{5}{6}$
$=\frac{11}{12}+\frac{1}{6}+\frac{5}{6}$
$=\frac{11}{12}+1=\frac{23}{12}$
o/
$\frac{20}{17}+\frac{-11}{2}-\frac{30}{4}$
$=\frac{20}{17}-\frac{11}{2}-\frac{15}{2}$
$=\frac{20}{7}-13=\frac{-71}{7}$
v/
$\frac{1}{2}-\frac{3}{4}+\frac{5}{6}$
$=\frac{6}{12}-\frac{9}{12}+\frac{10}{12}$
$=\frac{6-9+10}{12}=\frac{7}{12}$
3 tháng 7 2018
1) \(A=\frac{7}{10\times11}+\frac{7}{11\times12}+\frac{7}{12\times13}+...+\frac{7}{69\times70}\)
\(A=7\times\left(\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+...+\frac{1}{69\times70}\right)\)
\(A=7\times\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\times\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\times\frac{3}{35}\)
\(A=\frac{3}{5}\)
2) \(B=\frac{1}{25\times27}+\frac{1}{27\times29}+\frac{1}{29\times31}+...+\frac{1}{73\times75}\)
\(B=\frac{1}{2}\times\left(\frac{2}{25\times27}+\frac{2}{27\times29}+\frac{2}{29\times31}+...+\frac{2}{73\times75}\right)\).
\(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\times\frac{2}{75}\)
\(B=\frac{1}{75}\)
3) \(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)
\(C=\frac{4}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2008\times2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=2\times\frac{502}{1005}\)
\(C=\frac{1004}{1005}\)
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