tinh gia tri cua bieu thuc B={1(1/5).[4.(3+1/3-3/7-3/53)]/[3+1/3-3/7-3/53]}:{[4+4/17+4/19+4/2003]/5+5/17+519+5/2003]}
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U
10 tháng 8 2018
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
10 tháng 8 2018
\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
AK
31 tháng 5 2018
P/s : Đề của bạn sai nên mik đã sửa lại rồi
Ta có :
\(B=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(\Rightarrow B=-\frac{6}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{1\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(\Rightarrow B=-\frac{6}{5}.4:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}.\frac{5}{4}\)
\(\Rightarrow B=-6\)
Vậy \(B=-6\)
~ Ủng hộ nhé
HH
0
12 tháng 5 2022
Sửa đề; \(\dfrac{1}{5}\cdot\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{1}{5}\cdot4:\dfrac{4}{5}=\dfrac{4}{5}\cdot\dfrac{5}{4}=1\)
24 tháng 7 2015
A = \(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
A = \(-1\frac{1}{5}.\)4 : \(\frac{4.\left(1-\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5.\left(1-\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
A = \(-1\frac{1}{5}.4\): \(\frac{4}{5}\)= \(\frac{-6}{5}\).4. \(\frac{5}{4}\)
A = \(\frac{-24}{5}.\frac{5}{4}\)=\(\frac{\left(-6\right).1}{1.1}\)= -6.
HD
Hà Đức Thọ
Admin
VIP
24 tháng 7 2015
\(A=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(=-1\frac{1}{5}.\frac{4}{1}:\frac{4}{5}\)
\(=-1\frac{1}{5}.\frac{4}{1}.\frac{5}{4}\)
\(=-1\)
SS
2
DP
17 tháng 7 2017
B = 1.2.3 + 2.3.4 + 3.4.5 + .... + 17.18.19
<=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 17.18.19.4
<=> 4B = 1.2.3.(4 - 0) + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + .... + 17.18.19.(20 - 16)
<=> 4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .... + 17.18.19.20 - 16.17.18.19
<=> 4B = 17.18.19.20 = 116280
<=> B = 116280 : 4 = 29070
BB
17 tháng 7 2017
B = 1.2.3 + 2.3.4 + 3.4.5 + .... + 17.18.19
<=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 17.18.19.4
<=> 4B = 1.2.3.(4 - 0) + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + .... + 17.18.19.(20 - 16)
<=> 4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .... + 17.18.19.20 - 16.17.18.19
<=> 4B = 17.18.19.20 = 116280
<=> B = 116280 : 4 = 29070