Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
Bài 1:
\(a,\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2-\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{24+2-3}{12}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{46}.\dfrac{23}{12}\)
\(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{24}\)
\(x+\dfrac{1}{4}=\dfrac{7}{24}+\dfrac{1}{3}\)
\(x+\dfrac{1}{4}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}-\dfrac{1}{4}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
\(b,\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{1}{6}\)
\(\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\)
\(\dfrac{13}{21}+x=\dfrac{2}{7}\)
\(x=\dfrac{2}{7}-\dfrac{13}{21}=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
Bài 2:
\(a,\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(\dfrac{121}{12}-\dfrac{19}{2}\right)\)
\(=\dfrac{77}{18}:\dfrac{7}{12}\)
\(=\dfrac{22}{3}\)
\(b,1\dfrac{5}{18}-\dfrac{5}{18}.\left(\dfrac{1}{15}+1\dfrac{1}{12}\right)\)
\(=\dfrac{23}{18}-\dfrac{5}{18}.\dfrac{69}{60}\)
\(=\dfrac{23}{18}-\dfrac{23}{72}\)
\(=\dfrac{23}{24}\)
\(c,-\dfrac{1}{7}.\left(9\dfrac{1}{2}-8,75\right):\dfrac{2}{7}+0,625:1\dfrac{2}{3}\)
\(=\dfrac{-1}{7}.\dfrac{3}{4}:\dfrac{2}{7}+\dfrac{5}{8}:\dfrac{5}{3}\)
\(=-\dfrac{3}{8}+\dfrac{5}{8}:\dfrac{5}{3}\)
\(=-\dfrac{3}{8}+\dfrac{3}{8}\)
\(=\dfrac{0}{8}=0\)
Chúc bạn học tốt
ukm
bn có thể giải cho mik mấy bài mà mik vừa đăng đc ko mik đang cần gấp
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
a: \(=\left(\dfrac{19}{6}-\dfrac{2}{5}\right):\left(\dfrac{29}{6}+\dfrac{7}{10}\right)\)
\(=\dfrac{19\cdot5-2\cdot6}{30}:\dfrac{290+42}{30}=\dfrac{83}{332}=\dfrac{1}{4}\)
b: \(=\dfrac{\left(\dfrac{102}{25}-\dfrac{2}{25}\right)\cdot\dfrac{17}{4}}{\left(6+\dfrac{5}{9}-3-\dfrac{1}{4}\right)\cdot\dfrac{16}{7}}\)
\(=\dfrac{4\cdot\dfrac{17}{4}}{\dfrac{16}{7}\cdot\dfrac{119}{36}}=\dfrac{17}{\dfrac{68}{9}}=17\cdot\dfrac{9}{68}=\dfrac{9}{4}\)
c: \(=\left(\dfrac{120}{60}-\dfrac{15}{60}+\dfrac{20}{60}-\dfrac{36}{60}\right):\left(\dfrac{45}{15}-\dfrac{3}{15}-\dfrac{25}{15}\right)\)
\(=\dfrac{89}{60}:\dfrac{17}{15}=\dfrac{89}{60}\cdot\dfrac{15}{17}=\dfrac{89}{68}\)
9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)
\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)
10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)
b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)
c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)
d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)
\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\dfrac{-1}{4}\)
\(=\dfrac{7}{6}+\dfrac{-1}{4}\)
\(=\dfrac{11}{12}\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)
=>A>0(1)
\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)
=>B<0(2)
C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)
=>C=0(3)
Từ 1;2;3 =>A>C>B
\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)
\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)
\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)