Tính C= \(3\dfrac{1}{117}\cdot4\dfrac{1}{119}\cdot-1\dfrac{117}{116}\cdot5\dfrac{115}{119}+\dfrac{5}{119}-\dfrac{10}{117}\)
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18 tháng 6 2017
Đặt : \(\dfrac{1}{117}\) = x ; \(\dfrac{1}{119}\) = y .
A = ( 3 + x)( 4 + y) - (1 + 1 - x)(5 + 1 - y) - 5y
<=> A = 12 + 3y + 4x + xy - ( 2 - x)( 6 - y) - 5y
<=> A = 12 + 3y + 4x + xy - 12 + 2y + 6x - xy - 5y
<=> A = 10x
<=> A = \(\dfrac{10}{117}\).
Vậy A = \(\dfrac{10}{117}\)
25 tháng 6 2021
Ta có: \(A=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{352}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{352-2852-5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{-835}{4641}+\dfrac{8}{39}\)
\(=\dfrac{3}{119}\)
2 tháng 9 2022
Đặt 117=a; 119=b
Theo đề, ta có:
\(B=\left(3+\dfrac{1}{a}\right)\cdot\dfrac{1}{b}-\dfrac{4}{a}\cdot\left(5+\dfrac{b-1}{b}\right)-\dfrac{5}{a\cdot b}+8:\dfrac{a}{3}\)
\(=\dfrac{3a+1}{a}\cdot\dfrac{1}{b}-\dfrac{4}{a}\cdot\dfrac{5b+b-1}{b}-\dfrac{5}{ab}+\dfrac{24}{a}\)
\(=\dfrac{3a+1-24b+4-5}{ab}+\dfrac{24}{a}=\dfrac{3a-24b+24b}{ab}=\dfrac{3a}{ab}=\dfrac{3}{b}=\dfrac{3}{119}\)
4 tháng 9 2017
Đặt \(a=\dfrac{1}{117};b=\dfrac{1}{119}\)
Ta có:\(3\dfrac{1}{117}.\dfrac{1}{119}-\dfrac{4}{117}.5\dfrac{118}{119}-\dfrac{5}{117.119}+\dfrac{8}{39}\)
=\(\left(3+a\right)b-4a\left(6-b\right)-5ab+24a\)
\(\text{ }\)=3b+ab-24a+4ab-5ab+24a
\(=3b=3.\dfrac{1}{119}=\dfrac{3}{119}\)
Q
25 tháng 6 2017
\(N=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{1}{39}+\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{\dfrac{117119}{1000}}+\dfrac{8}{39}\)
\(=\dfrac{1}{4641}-\dfrac{2852}{13923}-\dfrac{5000}{117119}+\dfrac{8}{39}\)
\(=-\dfrac{9827881}{232949691}\)
9 tháng 7 2017
Đặt \(\dfrac{1}{117}=x;\dfrac{1}{119}=y\)
\(\Rightarrow\dfrac{1}{39}=3x\)
Ta có: \(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+8.3x\)
\(=3y+xy-20x-4x+4xy-5xy+24x\)
\(=3y\)
Thay \(y=\dfrac{1}{119}\rightarrow A:\)
\(A=3.\dfrac{1}{119}=\dfrac{3}{119}\)
Vậy \(A=\dfrac{3}{119}.\)
9 tháng 7 2017
Đặt \(a=\dfrac{1}{117};b=\dfrac{1}{119}\) thay vào A được:
A=\(\left(3+a\right)b-4a\left(6-b\right)-5ab+\dfrac{8}{39}\)
=\(3b+ab-24a+4ab-5ab+\dfrac{8}{39}\)
=\(3b-24a+\dfrac{8}{39}\) (1)
Thay \(a=\dfrac{1}{117};b=\dfrac{1}{119}\) vào (1) ta đuợc:
A=\(\dfrac{3}{119}-\dfrac{24}{117}+\dfrac{8}{39}=\dfrac{3}{119}-0=\dfrac{3}{119}\)
Chúc các bn học tốt
22 tháng 3 2018
\(tuA=1003+1007+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{2010}{119}=2010\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\)\(mauA=1003+1008+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{2011}{119}=2011\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\)có \(\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\ne0=>A=\dfrac{2010}{2011}\)
Sửa đề: \(C=3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}+\dfrac{5}{119}-\dfrac{10}{117}\)
\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(1+1-\dfrac{1}{117}\right)\left(5+1-\dfrac{1}{110}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)
\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(2-\dfrac{1}{117}\right)\left(6-\dfrac{1}{119}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)
Đặt \(a=\dfrac{1}{117}\) và \(b=\dfrac{1}{119}\) ta có:
\(C=\left(3+a\right).\left(4+b\right)-\left(2-a\right)\left(6-b\right)+5b-10a\)
\(=12+3b+4a+ab-12+2b+6a-ab+5b-10a\)
\(=10b=10.\dfrac{1}{119}=\dfrac{10}{119}\)
10\119