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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 = \(\dfrac{10}{117}\)
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
\(=3b=3.\dfrac{1}{119}=\dfrac{3}{119}\)
\(N=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{1}{39}+\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{\dfrac{117119}{1000}}+\dfrac{8}{39}\)
\(=\dfrac{1}{4641}-\dfrac{2852}{13923}-\dfrac{5000}{117119}+\dfrac{8}{39}\)
\(=-\dfrac{9827881}{232949691}\)
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 \(A=\dfrac{3}{119}.\)
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
a: Đặt 117=a; 119=b
\(A=3\frac{1}{117}\cdot4\frac{1}{119}-1\frac{116}{117}\cdot5\frac{118}{119}-\frac{5}{119}\)
\(=3\frac{1}{a}\cdot4\frac{1}{b}-\left(1+\frac{a-1}{a}\right)\cdot\left(5+\frac{b-1}{b}\right)-\frac{5}{b}\)
\(=\frac{3a+1}{a}\cdot\frac{4b+1}{b}-\frac{2a-1}{a}\cdot\frac{6b-1}{b}-\frac{5}{b}\)
\(=\frac{\left(3a+1\right)\left(4b+1\right)-\left(2a-1\right)\left(6b-1\right)-5a}{ab}\)
\(=\frac{12ab+3a+4b+1-\left(12ab-2a-6b+1\right)-5a}{ab}\)
\(=\frac{12ab+3a+4b+1-12ab+2a+6b-1-5a}{ab}=\frac{10b}{ab}=\frac{10}{a}\)
\(=\frac{10}{117}\)
b: Đặt 105=a; 651=b
\(B=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{3a}\right)\cdot\frac{1}{b}-\frac{1}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{3a\cdot b}+\frac{4}{a}\)
\(=\frac{6a+1}{3a}\cdot\frac{1}{b}-\frac{1}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{3ab}+\frac{4}{a}\)
\(=\frac{6a+1}{3ab}-\frac{4b-1}{ab}-\frac{4}{3ab}+\frac{4}{a}=\frac{6a+1-3\left(4b-1\right)-4+12b}{3ab}\)
\(=\frac{6a-3+12b-12b+3}{3ab}=\frac{6a}{3ab}=\frac{2}{b}=\frac{2}{651}\)
\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)
\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}-\dfrac{x+110}{117}-\dfrac{x+110}{119}=0\)
\(\Leftrightarrow\left(x+110\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)
Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)
\(\Rightarrow x+110=0\)
\(\Rightarrow x=-110\)
\(\dfrac{x-3}{133}+\dfrac{x-5}{155}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)
\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}=\dfrac{x+130}{117}+\dfrac{x+130}{119}\)
\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}-\dfrac{x+130}{117}-\dfrac{x+130}{119}=0\)
\(\Leftrightarrow\left(x+130\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)
Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)
\(\Leftrightarrow x+130=0\)
\(\Leftrightarrow x=-130\)
Vậy..
Ta có: \(A=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{352}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{352-2852-5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{-835}{4641}+\dfrac{8}{39}\)
\(=\dfrac{3}{119}\)
\(3\frac{1}{117}\)x \(4\frac{1}{119}\)- \(1\frac{116}{117}\)x \(5\frac{115}{119}\)- \(\frac{5}{119}\)
= \(\frac{1889}{13923}\)
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