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a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
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: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
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: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
a) \(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\Leftrightarrow\dfrac{3}{7}x-\dfrac{7}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{1}{2}+\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{17}{6}\)
\(\Leftrightarrow x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(\Leftrightarrow x=\dfrac{119}{18}\)
b) \(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{7}-\dfrac{2}{3}:x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{4}{7}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{56}{-57}\)
c) \(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\Leftrightarrow\left(\dfrac{2}{3}x+\dfrac{9}{4}\right):\dfrac{16}{6}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=\dfrac{3}{4}.\dfrac{16}{6}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=2\)
\(\Leftrightarrow\dfrac{2}{3}x=2-\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{-1}{4}\)
\(\Leftrightarrow x=\dfrac{-1}{4}:\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{-3}{8}\)
d) \(\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}-\dfrac{2}{3}=1\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=1+\dfrac{2}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{3}.\dfrac{1}{4}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4}{5}-\dfrac{2}{3}x=\dfrac{5}{12}\\\dfrac{4}{5}-\dfrac{2}{3}x=-\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{5}{12}\\\dfrac{2}{3}x=\dfrac{4}{5}-\left(-\dfrac{5}{12}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{23}{60}\\\dfrac{2}{3}x=\dfrac{73}{60}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{60}:\dfrac{2}{3}\\x=\dfrac{73}{60}:\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{73}{20}\end{matrix}\right.\)
a)\(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\dfrac{3}{7}x=0,5+2\dfrac{1}{3}\)
\(\dfrac{3}{7}x=\dfrac{17}{6}\)
\(x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(x=\dfrac{119}{18}\)
b)\(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=\dfrac{4}{7}-1\dfrac{1}{4}\)
\(x=\dfrac{2}{3}:\left(-\dfrac{19}{28}\right)\)
\(x=-\dfrac{56}{57}\)
c)\(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\dfrac{2}{3}x+2\dfrac{1}{4}=0,75:3\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{15}{64}-2\dfrac{1}{4}\)
\(x=-\dfrac{129}{64}:\dfrac{2}{3}\)
\(x=-\dfrac{387}{128}\)
a: Sửa đề: \(\left[\left(6\dfrac{3}{7}-\dfrac{0,75x-2}{0,35}\right)\cdot2.8+1.75\right]\cdot0.05=2.35\)
\(\Leftrightarrow\left(\dfrac{45}{7}-\dfrac{15x-40}{7}\right)\cdot2.8+1,75=47\)
\(\Leftrightarrow\dfrac{45-15x+40}{7}\cdot2,8=45,25\)
\(\Leftrightarrow\dfrac{85-15x}{7}=\dfrac{905}{56}\)
=>8(85-15x)=905
=>680-120x=905
=>120x=-225
hay x=-15/8
b: \(\dfrac{3}{5}-\dfrac{2}{7}< \dfrac{2}{3}x+\dfrac{3}{4}< \dfrac{1}{2}+\dfrac{7}{9}\)
\(\Leftrightarrow\dfrac{11}{35}-\dfrac{3}{4}< \dfrac{2}{3}x< \dfrac{23}{18}-\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{61}{140}< \dfrac{2}{3}x< \dfrac{19}{36}\)
\(\Leftrightarrow-\dfrac{183}{280}< x< \dfrac{19}{24}\)