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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!!!
mk có thắc mắc là bạn để 3 ra ngoài sao 1/3 vẫn giữ nguyên vậy phải bằng 1/9 mới đúng chứ'
Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé!
Lời giải:
a, \(\dfrac{12}{19}.\dfrac{7}{15}.\dfrac{-13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\dfrac{12}{19}.\dfrac{-7}{15}.\dfrac{13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\left(\dfrac{12}{19}.\dfrac{19}{12}\right).\left(\dfrac{13}{17}.\dfrac{17}{13}\right).\dfrac{-7}{15}\)
\(=1.1.\dfrac{-7}{15}\)
\(=\dfrac{-7}{15}\)
b, \(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{27}=\dfrac{-5}{9}\\\dfrac{-45}{b}=\dfrac{-5}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-5.27\\-5b=-45.9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-135\\-5b=-405\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-15\\b=-81\end{matrix}\right.\)
Vậy, \(a=-15;b=-81\).
\(C=\dfrac{4}{9}\times\dfrac{13}{17}+\dfrac{4}{17}\times\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{4}{9}\times\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\\ =\dfrac{4}{9}\times1+\dfrac{2}{9}\\ =\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{6}{9}=\dfrac{2}{3}\)
\(D=\dfrac{8}{19}\times\dfrac{5}{11}+\dfrac{7}{11}\times\dfrac{8}{19}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\dfrac{12}{11}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{12}{11}\times\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\\ =\dfrac{12}{11}\times19\\ =\dfrac{12}{11}\)
\(C=\dfrac{4}{9}\cdot\dfrac{13}{17}+\dfrac{4}{17}\cdot\dfrac{4}{9}+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}\cdot\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}\cdot\dfrac{13+4}{17}+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}\cdot\dfrac{17}{17}+\dfrac{9}{2}\)
\(C=\dfrac{4}{9}\cdot1+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}+\dfrac{2}{9}\)
\(C=\dfrac{4+2}{9}\)
\(C=\dfrac{6}{9}\)
\(C=\dfrac{2}{3}\)
\(D=\dfrac{8}{19}\cdot\dfrac{5}{11}+\dfrac{7}{11}\cdot\dfrac{8}{19}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)
\(D=\dfrac{8}{19}\cdot\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\cdot\dfrac{11}{19}\)
\(D=\dfrac{8}{19}\cdot\dfrac{12}{11}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)
\(D=\dfrac{12}{11}\cdot\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\)
\(D=\dfrac{12}{11}\cdot\dfrac{19}{19}\)
\(D=\dfrac{12}{11}\cdot1\)
\(D=\dfrac{12}{11}\)
`f)(-2)/17 + 15/23 + (-15)/17 + 4/19 + 8/23`
`= (-2/17+ -15/17)+(15/23+8/23)+4/19`
`= -1+1+4/19`
`= 0 +4/19`
`= 0`
`g)(-1)/2 + 3/21 + (-2)/6 + (-5)/30`
`= (-1)/2 + 1/7 + (-1)/3 + (-1)/6`
`= (-21)/42 + 6/42 + (-14)/42 + (-7)/42`
`=(-36)/42`
`=(-6)/7`
f)\(-\dfrac{2}{17}+\dfrac{15}{23}+-\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
\(=\left(-\dfrac{2}{17}+-\dfrac{15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\)
\(=-1+1+\dfrac{4}{19}\)
\(=0+\dfrac{4}{19}=\dfrac{4}{19}\)
g)\(-\dfrac{1}{2}+\dfrac{3}{21}+-\dfrac{2}{6}+-\dfrac{5}{30}\)
\(=-\dfrac{1}{2}+\dfrac{1}{7}+-\dfrac{1}{3}+-\dfrac{1}{6}\)
\(=\left(-\dfrac{1}{2}+-\dfrac{1}{3}+-\dfrac{1}{6}\right)+\dfrac{1}{7}\)
\(=-\dfrac{3+2+1}{6}+\dfrac{1}{7}\)
\(=\dfrac{1}{7}-1\)
\(=\dfrac{1}{7}-\dfrac{7}{7}=-\dfrac{6}{7}\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.4}=-6\)
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
`B=11/17+ (-8/19) +(-3/4) + 6/17 - 30/19`
`B=(11/17 +6/17) + (-8/19 -30/19) + (-3/4)`
`B= 17/17+(-38/19)+(-3/4)`
`B= 1 +(-2) +(-3/4)`
`B= -1 +(-3/4)`
`B=-7/4`