a) \(\dfrac{1}{2}+\dfrac{-1}{-3}-\dfrac{5}{12}< 2x< \dfrac{12}{-31}+\dfrac{136}{31}\)
b) \(\dfrac{-2}{5}< \dfrac{x}{15}< \dfrac{1}{6}\)
HéP :)
Những câu hỏi liên quan
A=\(\left[\dfrac{1\dfrac{11}{31}.4\dfrac{3}{7}-\left(15-6\dfrac{1}{3}.\dfrac{2}{19}\right)}{4\dfrac{5}{6}+\dfrac{1}{6}\left(12-5\dfrac{1}{3}\right)}.\left(-1\dfrac{14}{93}\right)\right].\dfrac{31}{50}\)
tính: A=\([\dfrac{1\dfrac{11}{31}.4\dfrac{3}{7}-(15-6\dfrac{1}{3}.\dfrac{2}{19})}{4\dfrac{5}{6}+\dfrac{1}{6}(12-5\dfrac{1}{3})}.(-1\dfrac{14}{93})].\dfrac{31}{50}\)
kết quả là 1 nha, bn có thể tính nhẩm sẽ ra nhanh thôi
( chúc bạn học tốt)
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\(\dfrac{2}{5}\)*\(15\dfrac{1}{3}\)-\(\dfrac{2}{5}\)*\(10\dfrac{1}{3}\)
\(12\dfrac{5}{11}\)-(\(3\dfrac{1}{4}\)+\(2\dfrac{5}{11}\)]
\(\dfrac{34}{31}\)-\(\dfrac{19}{28}\)-\(\dfrac{3}{31}\)
\(\dfrac{2}{5}\times15\dfrac{1}{3}-\dfrac{2}{5}\times10\dfrac{1}{3}\)
\(=\dfrac{2}{5}\times\left(15\dfrac{1}{3}-10\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\times5\)
\(=2\)
____________________
\(12\dfrac{5}{11}-\left(3\dfrac{1}{4}+2\dfrac{5}{11}\right)\)
\(=12\dfrac{5}{11}-3\dfrac{1}{4}-2\dfrac{5}{11}\)
\(=\left(12\dfrac{5}{11}-2\dfrac{5}{11}\right)-3\dfrac{1}{4}\)
\(=10-3\dfrac{1}{4}\)
\(=\dfrac{27}{4}\)
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\(\dfrac{34}{31}-\dfrac{19}{28}-\dfrac{3}{31}\)
\(=\left(\dfrac{34}{31}-\dfrac{3}{31}\right)-\dfrac{19}{28}\)
\(=\dfrac{31}{31}-\dfrac{19}{28}\)
\(=1-\dfrac{19}{28}\)
\(=\dfrac{9}{28}\)
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\(a.19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12} b.\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
c.\(\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-\left(4\dfrac{1}{5}\right)\right)-\dfrac{11}{31}\)
1.Tính giái trị biểu thức
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
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Tính giá trị biểu thức :
A=\(\left[\dfrac{1\dfrac{11}{31}.4\dfrac{3}{7}-\left(15-6\dfrac{1}{3}.\dfrac{2}{19}\right)}{4\dfrac{5}{6}+\dfrac{1}{6}\left(12-5\dfrac{1}{3}\right)}\right].\dfrac{31}{50}\)
A=\(\left[\dfrac{\dfrac{42}{31}.\dfrac{31}{7}-\left(15-\dfrac{2}{3}\right)}{\dfrac{29}{6}+\dfrac{1}{6}.\dfrac{20}{3}}\right].\dfrac{31}{50}\)
= \(\left(\dfrac{6-\dfrac{43}{3}}{\dfrac{29}{6}+\dfrac{10}{9}}\right).\dfrac{31}{50}\)=\(\left(\dfrac{\dfrac{-25}{3}}{\dfrac{107}{18}}\right).\dfrac{31}{50}\)=\(\dfrac{-150}{107}.\dfrac{31}{50}\)=\(\dfrac{-93}{107}\)
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Điền dấu < ; > ; =
3\(\dfrac{1}{5}-\dfrac{8}{3}...........\dfrac{17}{5}-2\dfrac{12}{5}\)
\(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{89}+\dfrac{1}{90}\) ..... \(\dfrac{5}{6}\)
27 tháng 3 2022 lúc 7:51
Tìm x, biết \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36
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Bài 3:a) dfrac{2x-1}{5}-dfrac{x-2}{3}dfrac{x+7}{15}b) dfrac{x+3}{2}-dfrac{x-1}{3}dfrac{x+5}{6}+1c) dfrac{2left(x+5right)}{3}+dfrac{x+12}{2}-dfrac{5left(x-2right)}{6}dfrac{x}{3}+11d) dfrac{x-4}{5}+dfrac{3x-2}{10}-xdfrac{2x-5}{3}-dfrac{7x+2}{6}e) dfrac{left(2x-3right)left(2x+3right)}{8}dfrac{left(x-4^{ }right)^2}{6}+dfrac{left(x-2right)^2}{3}d) dfrac{7x^2-14x-5}{15}dfrac{left(2x+1right)^2}{5}-dfrac{left(x-1right)^2}{3}e) dfrac{left(7x+1right)left(x-2right)}{10}+dfrac{2}{5}dfrac{left(x-2right)^2}{5...
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Bài 3:
a) \(\dfrac{2x-1}{5}\)-\(\dfrac{x-2}{3}\)
=\(\dfrac{x+7}{15}\)
b) \(\dfrac{x+3}{2}\)-\(\dfrac{x-1}{3}\)
=\(\dfrac{x+5}{6}\)+1
c) \(\dfrac{2\left(x+5\right)}{3}\)+\(\dfrac{x+12}{2}\)
-\(\dfrac{5\left(x-2\right)}{6}\)=\(\dfrac{x}{3}\)+11
d) \(\dfrac{x-4}{5}\)+\(\dfrac{3x-2}{10}\)-x
=\(\dfrac{2x-5}{3}\)-\(\dfrac{7x+2}{6}\)
e) \(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\)
=\(\dfrac{\left(x-4^{ }\right)^2}{6}\)+\(\dfrac{\left(x-2\right)^2}{3}\)
d) \(\dfrac{7x^2-14x-5}{15}\)
=\(\dfrac{\left(2x+1\right)^2}{5}\)-\(\dfrac{\left(x-1\right)^2}{3}\)
e) \(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}\)+\(\dfrac{2}{5}\)
=\(\dfrac{\left(x-2\right)^2}{5}\)+\(\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)
a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
\(\Leftrightarrow6x-3-5x+10-x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
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Tìm x biết :
a)\(\dfrac{x+1}{2x+1}=\dfrac{\dfrac{1}{2}x+2}{x+3}\)
b) \(x^2+5x=0\)
c)\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)
=>(x+1)(2x+6)=(2x+1)(x+4)
\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)
=>9x+4=8x+6
=>x=2
b: \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
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