(5+10+15+...+1000)\(\left[\dfrac{2}{5}:0.5+2\left(-0.4\right)\right]:\left(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{15}+...\dfrac{1}{1000}\right)\)
Những câu hỏi liên quan
1. 25dfrac{1}{7}:left(-dfrac{5}{7}right)-15dfrac{1}{7}:left(-dfrac{5}{7}right)+dfrac{4}{5} 3. 2dfrac{2}{3}:left{left[left(3,72-0.02right)dfrac{10}{37}right]:dfrac{5}{6}+2,8right}-dfrac{7}{15}2. left(3+dfrac{4}{5}-dfrac{5}{12}right)left(dfrac{6}{7}-dfrac{3}{5}right)^24.23+3.left(-dfrac{1}{2}right)^2-22.4+left[left(-2right)^2:dfrac{1}{2}right]
Đọc tiếp
1. \(25\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)-15\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)+\dfrac{4}{5}\) 3. \(2\dfrac{2}{3}:\left\{\left[\left(3,72-0.02\right)\dfrac{10}{37}\right]:\dfrac{5}{6}+2,8\right\}-\dfrac{7}{15}\)
2. \(\left(3+\dfrac{4}{5}-\dfrac{5}{12}\right)\left(\dfrac{6}{7}-\dfrac{3}{5}\right)^2\)
4.23+3.\(\left(-\dfrac{1}{2}\right)^2\)-22.4+\(\left[\left(-2\right)^2:\dfrac{1}{2}\right]\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
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\(\dfrac{\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\right):\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{1}{15}\right)}{\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\right):\left(\dfrac{1}{4}-\dfrac{1}{6}\right)}=\)
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
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=\(\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
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Giải các phương trình sau:
\(h.\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
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\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)
\(\Leftrightarrow15x+9x-1=14-7x\)
\(\Leftrightarrow31x=15\)
\(\Leftrightarrow x=\dfrac{15}{31}\)
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Tính:
a) \(\dfrac{5}{16}-\dfrac{5}{24};\)
b) \(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)\);
c) \(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right).\)
\(a.\)
\(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{5\cdot3-5\cdot2}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
\(b.\)
\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2-5-9}{11}=-\dfrac{12}{11}\)
\(c.\)
\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{5}{12}+\dfrac{1}{15}=\dfrac{6-5\cdot5+4}{60}=-\dfrac{15}{60}=-\dfrac{1}{4}\)
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a) \(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
b)\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2}{11}-\dfrac{5}{11}-\dfrac{9}{11}=\dfrac{2-5-9}{11}=\dfrac{-12}{11}\)
c)\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{7}{20}=\dfrac{2}{20}-\dfrac{7}{20}=\dfrac{-5}{20}=\dfrac{-1}{4}\)
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23 tháng 1 lúc 19:43
Tìm \(x\) biết:
\(\left(\sqrt{3}\right)^x=243\)
\(0,1^x=1000\)
\(\left(\dfrac{1}{2}\right)^x=1024\)
\(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
\(5^{x-1}+5^{x+2}=3\)
a: \(\left(\sqrt{3}\right)^x=243\)
=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)
=>\(\dfrac{1}{2}\cdot x=5\)
=>x=10
b: \(0,1^x=1000\)
=>\(\left(\dfrac{1}{10}\right)^x=1000\)
=>\(10^{-x}=10^3\)
=>-x=3
=>x=-3
c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
=>\(\left(0,2\right)^{x+3}< 0,2\)
=>x+3>1
=>x>-2
d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)
=>2x+1<-2
=>2x<-3
=>\(x< -\dfrac{3}{2}\)
e: \(5^{x-1}+5^{x+2}=3\)
=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)
=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)
=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)
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1. dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4}dfrac{2left(2x+1right)}{7}-5
2. x-dfrac{3left(x+30right)}{15}-24dfrac{1}{2}dfrac{7x}{10}-dfrac{2left(10x+2right)}{5}
3. 14dfrac{1}{2}-dfrac{2left(x+3right)}{5}dfrac{3x}{2}-dfrac{2left(x-7right)}{3}
4. dfrac{x+1}{3}+dfrac{3left(2x+1right)}{4}dfrac{2x+3left(x+1right)}{6}+dfrac{7+12x}{12}
5. dfrac{3left(2x-1right)}{4}-dfrac{3x+1}{10}+1dfrac{2left(3x+2right)}{5}
6. x-dfrac{3}{17}left(2x-1right)dfrac{7}{34}left(1-2xright)+dfrac{10x-3}{2}
7. dfrac{3le...
Đọc tiếp
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
Thực hiện các phép tính:
a,\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
b, \(10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+3:15\%\)
c. \(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,8\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
a, = (58/9 + 7/11) - (40/9 - 26/11)
= 701/99 - 206/99
= 5
b, = 51/5 - 11/2 . 60/11 + 3 : 3/20
= 51/5 - 30 + 20
= -99/5 + 20
= 1/5
c, = 19/4 + (-0,37) + 1/8 + (-1,8) + (-2,5) + 37/12
= 219/50 + -67/40 + 7/12
= 1973/600
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tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
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b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
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c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))
C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))
C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\) - \(\dfrac{588}{105}\)]
C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]
C = - 15.(- \(\dfrac{143}{105}\))
C = \(\dfrac{143}{7}\)
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dfrac{8^{14}}{4^4.64^5}
dfrac{9^{10}.27^7}{81^7.3^{15}}
left(dfrac{3}{10}right)^4.left(0,3right)^5.left(dfrac{10}{3}right)^{10}
dfrac{left(4^3right)^2.9^4}{6^7.8^2}
dfrac{4^8.9^4}{6^6.8^3}
3^6.left(dfrac{1}{3}right)^6.81^2.dfrac{1}{27^2} TÍNH
Đọc tiếp
\(\dfrac{8^{14}}{4^4.64^5}\)
\(\dfrac{9^{10}.27^7}{81^7.3^{15}}\)
\(\left(\dfrac{3}{10}\right)^4.\left(0,3\right)^5.\left(\dfrac{10}{3}\right)^{10}\)
\(\dfrac{\left(4^3\right)^2.9^4}{6^7.8^2}\)
\(\dfrac{4^8.9^4}{6^6.8^3}\)
\(3^6.\left(\dfrac{1}{3}\right)^6.81^2.\dfrac{1}{27^2}\) TÍNH
\(\dfrac{8^{14}}{4^4.64^5}=\dfrac{\left(2^3\right)^{14}}{\left(2^2\right)^4.\left(2^5\right)^5}=\dfrac{2^{42}}{2^8.2^{25}}=2^{42-\left(8+25\right)}=2^9\)
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\(\dfrac{9^{10}.27^7}{81^7.3^{15}}=\dfrac{\left(3^2\right)^{10}.\left(3^3\right)^7}{\left(3^4\right)^7.3^{15}}=\dfrac{3^{20}.3^{21}}{3^{28}.3^{15}}=\dfrac{3^{20+21}}{3^{28+15}}=\dfrac{3^{41}}{3^{41}.3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)
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