(7,25x+11/9)÷(-2/3)^2=-3/5
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\(7,25x+\dfrac{11}{9}=-\dfrac{3}{5}.-\dfrac{27}{8}\\ 7,25x+\dfrac{11}{9}=\dfrac{81}{40}\\ 7,25x=\dfrac{289}{360}\\ x=\dfrac{289}{360}:7,25\\ x=\dfrac{289}{2610}.\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
Bài 1: Tính nhanh:
A = 3/1*2 + 3/2*3 + 3/3*4 + ... + 3/399*400
=>3A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/399*400
3A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/399 - 1/400
3A = 1 - 1/400
3A = 400/400 - 1/400
3A = 399/400
A = 399/400 : 3
A = 399/400 . 1/3
A = 133/400.
Có gì ko hiểu bn ib mk nha.^^
\(A=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{399.400}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)
\(A=3.\left(1-\frac{1}{400}\right)\)
\(A=3.\frac{399}{400}\)
\(A=\frac{1197}{400}\)
\(B=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{399.400}\)
\(B=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)
\(B=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)
\(B=5.\left(1-\frac{1}{400}\right)\)
\(B=5.\frac{399}{400}\)
\(B=\frac{399}{80}\)
\(C=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\)
\(C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\)
\(C=\frac{1}{5}-\frac{1}{151}\)
\(C=\frac{146}{755}\)
\(D=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{149.151}\)
\(D=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\right)\)
\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\right)\)
\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{151}\right)\)
\(D=\frac{3}{2}.\frac{146}{755}\)
\(D=\frac{219}{755}\)
\(E=\frac{11}{1.3}+\frac{11}{3.5}+\frac{11}{5.7}+...+\frac{11}{99.101}\)
\(E=\frac{11}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(E=\frac{11}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(E=\frac{11}{2}.\left(1-\frac{1}{101}\right)\)
\(E=\frac{11}{2}.\frac{100}{101}\)
\(E=\frac{550}{101}\)
_Chúc bạn học tốt_
a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)
\(=\dfrac{5}{7}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{3}{5}\)
`=`\(\dfrac{9}{12}+\dfrac{8}{12}+\dfrac{3}{5}\)
`=`\(\dfrac{17}{12}+\dfrac{3}{5}\)
`=`\(\dfrac{85}{60}+\dfrac{36}{60}\)
`=`\(\dfrac{121}{60}\)
`b)`
\(\dfrac{1}{2}\cdot\dfrac{9}{13}\div\dfrac{27}{26}\)
`=`\(\dfrac{1}{2}\cdot\dfrac{9}{13}\cdot\dfrac{26}{27}\)
`=`\(\dfrac{1}{2}\cdot\dfrac{2}{3}\)
`=`\(\dfrac{1}{3}\)
`c)`
\(\dfrac{2}{7}\cdot\dfrac{1}{9}+\dfrac{2}{7}\cdot\dfrac{2}{9}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{2}{7}\cdot\left(\dfrac{1}{9}+\dfrac{2}{9}\right)+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{2}{7}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{1}{3}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
`=`\(\dfrac{1}{3}\cdot1=\dfrac{1}{3}\)
`d)`
\(11\div\dfrac{5}{2}+11\div\dfrac{7}{3}+11\div\dfrac{35}{6}\)
`=`\(11\cdot\dfrac{2}{5}+11\cdot\dfrac{3}{7}+11\cdot\dfrac{6}{35}\)
`=`\(11\cdot\left(\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{6}{35}\right)\)
`=`\(11\cdot1=11\)
a) 3/4 + 2/3 + 3/5 = 45/60 + 40/60 + 36/60 = 121/60
b) 1/2 x 9/13 : 27/26 = 9/26 x 26/27 = 1/3
c) 2/7 x 1/9 + 2/7 x 2/9 + 1/3 x 5/7 = 2/7 x (1/9 + 2/9) + 5/21 = 2/7 x 1/3 + 5/21 = 2/21 + 5/21 = 1/3
d) 11 : 5/2 + 11 : 7:3 + 11 : 35/6 = 11 x (2/5 + 3/7 + 6/35) = 11 x 1 = 11
a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)
\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)
\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)
=0
b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\cdot0=0\)
c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)
\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)
d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)
\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)
\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)
\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)
\(=0:\dfrac{3}{8}=0\)
\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)
\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)
\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)
\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)
\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)
\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)
\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)
\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
\(\left(7,25x+\frac{11}{9}\right)\div\left(\frac{-2}{3}\right)^2=\frac{-3}{5}\)
\(\left(\frac{29}{4}x+\frac{11}{9}\right)\div\frac{4}{9}=\frac{-3}{5}\)
\(\frac{29}{4}x+\frac{11}{9}=\frac{-3}{5}.\frac{4}{9}\)
\(\frac{29}{4}x+\frac{11}{9}=\frac{-4}{15}\)
\(\frac{29}{4}x=\frac{-4}{15}-\frac{11}{9}\)
\(\frac{29}{4}x=\frac{-67}{45}\)
\(x=\frac{-67}{45}\div\frac{29}{4}\)
\(x=\frac{-268}{1305}\)