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Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
\(N=8\dfrac{1}{5}\left(11\dfrac{94}{1591}-6\dfrac{38}{1517}\right):8\dfrac{11}{43}\)
\(N=\dfrac{41}{5}\left(\dfrac{17595}{1591}-\dfrac{9140}{1517}\right):\dfrac{355}{43}\)
\(N=\dfrac{41}{5}.\dfrac{8875}{1763}:\dfrac{355}{43}\)
\(N=\dfrac{1775}{43}:\dfrac{355}{43}\)
\(N=5.\)
Trời ơi cái đề bài !!!
Thoy thì làm từng câu vậy
a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\dfrac{7}{222222}\)
\(I=\dfrac{7}{22}\)
Bài 1 :
\(=\dfrac{2}{11}+\dfrac{4}{11}-\dfrac{6}{11}-\dfrac{3}{8}-\dfrac{5}{8}=0-1=-1\)
Bài 2 :
\(\Rightarrow3+x=8\Leftrightarrow x=5\)
Bài 3 :
\(\Leftrightarrow x-\dfrac{5}{11}=\dfrac{5}{4}\Leftrightarrow x=\dfrac{35}{44}\)
Bài 4 :
Trong 2 ngày An đọc được số quyên phần quyên sách
\(\dfrac{1}{11}+\dfrac{8}{11}=\dfrac{9}{11}\)( quyển sách )
đs : 9/11 quyển sách
\(M=\dfrac{120-\dfrac{1}{2}.40.5.\dfrac{1}{5}.20.\dfrac{1}{4}-20}{1+5+....+41}\\ =\dfrac{120-20.5-20}{1+5+...+41}\\ =\dfrac{0}{1+5+...+41}\\ =0\)
\(N=10101\left(\dfrac{6}{3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}-\dfrac{7}{3.7.11.13.37}\right)\\ =10101\left(-\dfrac{2}{2.3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}\right)\\ =3.7.13.37\left(\dfrac{4}{2.3.7.11.13.37}\right)=\dfrac{4}{2.11}=\dfrac{2}{11}\)
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) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
\(10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)=\(10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
= \(10101\cdot\left(\dfrac{5}{111111}-\dfrac{4}{111111}+\dfrac{5}{222222}\right)\)
= \(10101\cdot\left(\dfrac{1}{111111}+\dfrac{5}{222222}\right)\)
= \(10101\cdot\left(\dfrac{2}{222222}+\dfrac{5}{222222}\right)\)
= \(10101\cdot\dfrac{7}{222222}\)
= \(\dfrac{10101\cdot7}{222222}=\dfrac{10101\cdot7}{10101\cdot22}=\dfrac{7}{22}\)
10101.(\(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\))
= 10101.(\(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\))
= 10101.(\(\dfrac{1}{111111}+\dfrac{5}{222222}\))
= 10101. \(\dfrac{7}{222222}\)
= \(\dfrac{10101.7}{222222}\)= \(\dfrac{70707}{22222}\)=\(\dfrac{7}{22}\)
Chúc pạn hok tốt!!!!!!!
\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)
\(\dfrac{3}{2}x-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\dfrac{3}{2}x=\dfrac{3}{5}+\dfrac{1}{5}\)
\(\dfrac{3}{2}x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}:\dfrac{3}{2}\)
\(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{3}\)
\(x=\dfrac{9}{12}-\dfrac{4}{12}\)
\(x=\dfrac{5}{12}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x=\dfrac{1}{4}+\dfrac{2}{5}\)
\(\dfrac{3}{2}x=\dfrac{13}{20}\)
\(x=\dfrac{13}{20}:\dfrac{3}{2}\)
\(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)
\(x=\dfrac{13}{30}\)
\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}\)
\(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)
\(x=\dfrac{10}{3}\)
1. \(\dfrac{16.17-5}{16.16-1}:\left(x-\dfrac{2}{3}\right)=\dfrac{1}{3}\)
<=> \(\dfrac{272-5}{256-1}:\left(x-\dfrac{2}{3}\right)=\dfrac{1}{3}\)
<=> \(\dfrac{267}{255}:\left(x-\dfrac{2}{3}\right)=\dfrac{1}{3}\)
<=> x - \(\dfrac{2}{3}=\dfrac{267}{85}\)
<=> x = \(\dfrac{971}{255}\)
@Triều Nguyễn Quốc
2. a, 8\(\dfrac{1}{5}\left(11\dfrac{94}{1591}-6\dfrac{38}{1517}\right):8\dfrac{11}{13}\)
= \(\dfrac{41}{5}\left(5\dfrac{60}{1763}\right):\dfrac{355}{43}\)
= \(\dfrac{41}{5}.\dfrac{8875}{1763}:\dfrac{355}{43}\)
= \(\dfrac{1775}{43}:\dfrac{355}{43}\)
= 5
@Triều Nguyễn Quốc