\(\dfrac{\left(\dfrac{1}{2}\right)^{10}.5-\left(\dfrac{1}{4}\right)^5.3}{\dfrac{1}{1024}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
tính giá trị của biểu thức : B\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}.5-\left(\dfrac{1}{4}\right)^5.3}{\dfrac{1}{1024}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)
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
Thu gọn:
A = \(\dfrac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\) D = \(\dfrac{\left(\dfrac{1}{2}-\dfrac{1}{3}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}\right).\left(\dfrac{2}{3}+\dfrac{3}{4}\right)}\)
B = \(\dfrac{2^3-3^4-2^4.3^3}{2^5.3^4-2^6.3^3}\) E = \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{6}\right)\)
C = \(\dfrac{\dfrac{1}{2}-\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{3}{4}}{\dfrac{2}{3}-\dfrac{2}{3}:\dfrac{5}{6}-\dfrac{5}{6}}\)
Giúp mình với ! 10k nha
\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
\(3\dfrac{3}{3}.\dfrac{1}{3}-\dfrac{3}{4}.\dfrac{1}{3}\)
\(\left[\dfrac{11}{3}\right]-\left(\dfrac{-1}{2}\right)^2-4\dfrac{1}{2}\)
\(\left(\dfrac{3}{2}-\dfrac{5}{4}+\dfrac{1}{3}\right):\left(\dfrac{4}{3}+2\dfrac{3}{2}-\dfrac{3}{4}\right)\)
\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5+\dfrac{-5}{27}+\dfrac{16}{23}\)
\(2\dfrac{5}{4}+\left(-2018\right)^0-\left[\dfrac{-1}{4}\right]\)
\(\dfrac{19}{11}.\dfrac{6}{5}+\dfrac{6^2}{11}.\dfrac{6}{5}-\left(\dfrac{1}{2}\right)^0\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
a)\(2\dfrac{3}{3}.4.\left(-0,4\right)+1\dfrac{3}{5}.1,75+\left(-7,2\right):\dfrac{9}{11}\)
b)\(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}.\left(-5\right)^0\)
a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)
\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)
\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)
\(=\dfrac{-54}{5}\)
b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)
\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)
\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)
\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)
Kết quả học tập kì I của lớp 6A xếp thành 3 loại:giỏi,khá,trung bình.Số học sinh giỏi chiếm \(\dfrac{1}{3}\) số học sinh của lớp,số học sinh khá chiếm 40% số học sinh cả lớp,số học sinh trung bình là 12 em.Tính số học sinh lớp 6A và tỉ số phần trăm của học sinh giỏi so với học sinh cả lớp.
Thực hiện phép tính:
a) \(\left(\dfrac{-3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(\dfrac{-4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
b)\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
a: =11/7(-3/7+4/11-4/7+7/11)=0
b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)
Tính nhanh
A=\(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{1024}\right)\)
B=4.5100 .\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\right)+1\)