Nếu tan \(\dfrac{x}{2}=2\) thì giá trị của biểu thức \(\dfrac{sinx}{3-2cosx+5tanx}\) bằng
A. \(\dfrac{12}{37}\)
B. \(-\dfrac{12}{37}\)
C. \(\dfrac{11}{37}\)
D. \(-\dfrac{11}{37}\)
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Tìm giá trị lớn nhất hoặc nhỏ nhất của các đa thức sau:A dfrac{37}{x^2-2x+3} B dfrac{-26}{x^2-5x+10} C dfrac{-2023}{x^2-x+6} D dfrac{0,75}{x^2+x+5} E dfrac{13}{2x^2-x+37} F dfrac{-61}{3x^2-x+19}
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Tìm giá trị lớn nhất hoặc nhỏ nhất của các đa thức sau:
A =\(\dfrac{37}{x^2-2x+3}\) B \(=\dfrac{-26}{x^2-5x+10}\) C \(=\dfrac{-2023}{x^2-x+6}\) D \(=\dfrac{0,75}{x^2+x+5}\) E \(=\dfrac{13}{2x^2-x+37}\) F \(=\dfrac{-61}{3x^2-x+19}\)
\(a,\left(\dfrac{37}{9}+\dfrac{13}{4}\right)x\dfrac{9}{4}+\dfrac{11}{4}\) b,\(1+\left(\dfrac{9}{10}-\dfrac{-4}{5}\right):\dfrac{19}{6}\)
c,\(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}x\dfrac{12}{5}\)
Giúp mik nha:>
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
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28 tháng 1 2022 lúc 21:48
Tính giá trị biểu thức:e,dfrac{18}{37}+dfrac{8}{24}+dfrac{19}{37}-1dfrac{23}{24}+dfrac{2}{3}f,left(-2right)^3.left(dfrac{3}{4}-0,25right):left(2dfrac{1}{4}-1dfrac{1}{6}right)g,left(dfrac{2}{5}right)^2+5dfrac{1}{2}.left(4,5-2right)+dfrac{2^3}{left(-4right)}h,dfrac{4}{9}.19dfrac{1}{3}-dfrac{4}{9}.39dfrac{1}{3}i,left(-dfrac{1}{2}right)^2:dfrac{1}{4}-2left(-dfrac{1}{2}right)^2j,125%.left(dfrac{-1}{2}right)^2:left(1dfrac{5}{16}-1,5right)+2008^0k,left(-2right)^3.dfrac{-1}{24}+left(dfrac{4}{3}-1dfrac{5...
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Tính giá trị biểu thức:
\(e,\dfrac{18}{37}+\dfrac{8}{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac{2}{3}\)
\(f,\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(g,\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}.\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(h,\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\)
\(i,\left(-\dfrac{1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(j,125\%.\left(\dfrac{-1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2008^0\)
\(k,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
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Bài 2: Tính hợp lý:
\(A=\dfrac{63636337-37373763}{1+2+3+...+2006}\)
\(B=1\dfrac{6}{41}\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\right)\dfrac{124242423}{237373735}\)
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
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\(-\dfrac{5}{12}+\dfrac{4}{37}+\dfrac{7}{12}-\dfrac{41}{37}+\dfrac{1}{2}\)
\(\dfrac{-5}{12}+\dfrac{4}{37}+\dfrac{7}{12}-\dfrac{41}{37}+\dfrac{1}{2}\)
\(=\left(\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{1}{2}\right)+\left(\dfrac{4}{37}-\dfrac{41}{37}\right)\)
\(=\dfrac{2}{3}-1=-\dfrac{1}{3}\)
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23 tháng 3 2022 lúc 20:53
Giá trị của tích dfrac{1}{11} . dfrac{1}{12}bằng giá trị của biểu thức nào sau đây?A.dfrac{1}{12}-dfrac{1}{11} B.dfrac{1}{23} C.dfrac{1}{11}+dfrac{1}{12} D.dfrac{1}{11}-dfrac{1}{12}
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Giá trị của tích \(\dfrac{1}{11}\) . \(\dfrac{1}{12}\)bằng giá trị của biểu thức nào sau đây?
A.\(\dfrac{1}{12}\)-\(\dfrac{1}{11}\) B.\(\dfrac{1}{23}\) C.\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\) D.\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)
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Tính bằng cách hợp lí: \(B=1\dfrac{6}{41}\cdot\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\right)\cdot\dfrac{124242423}{237373735}\)
Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )
B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]
B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]
B = \(\dfrac{47}{41}.5\)
B = \(\dfrac{235}{41}\)
Chúc bn hc tốt!!!
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\(\dfrac{3}{5}\) x \(x\) \(-\) \(\dfrac{3}{2}\) = \(\dfrac{5}{6}\)
\(\dfrac{12}{15}+\dfrac{37}{7}+\dfrac{16}{5}+57\)
a: =>3/5x=5/6+3/2=5/6+9/6=14/6=7/3
=>x=35/9
b: =(4/5+16/5)+(37/7+57)
=61+37/7=464/7
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\(\dfrac{3}{5}x=\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{14}{6}=\dfrac{7}{3}\Leftrightarrow x=\dfrac{7}{3}:\dfrac{3}{5}=\dfrac{35}{9}\)
\(\dfrac{19}{5}+\dfrac{37}{7}+57=\dfrac{19.7+37.5}{35}+\dfrac{57.35}{35}=\dfrac{2313}{35}\)
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\(\dfrac{3}{5}\times x-\dfrac{3}{2}=\dfrac{5}{6}\)
\(\dfrac{3}{5}\times x=\dfrac{5}{6}+\dfrac{3}{2}\)
\(\dfrac{3}{5}\times x=\dfrac{7}{3}\)
\(x=\dfrac{7}{3}:\dfrac{3}{5}\)
\(x=\dfrac{7}{3}\times\dfrac{5}{3}\)
\(x=\dfrac{35}{9}\)
\(\dfrac{12}{15}+\dfrac{37}{7}+\dfrac{16}{5}+57=\dfrac{4}{5}+\dfrac{37}{3}+\dfrac{16}{5}+57=\left(\dfrac{4}{5}+\dfrac{16}{5}\right)+\left(\dfrac{37}{7}+57\right)=4+\dfrac{436}{7}=\dfrac{464}{7}\)
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Tính \(S=1\dfrac{6}{41}.\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4}{5}\right).\dfrac{124242423}{237373735}\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.4}=-6\)
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\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)
Vậy...
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