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\(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}=\frac{31}{1.3}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}+\frac{31}{11.13}\\ \)
\(=\frac{31}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{13}\right)=\frac{31}{2}.\frac{12}{13}=\frac{31.6}{13}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\Leftrightarrow x=\frac{195}{13}=15\)
\(x-\left(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\)\(\frac{9}{13}\)(1)
Đặt \(A=\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\)
\(A=31\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(\Rightarrow2A=31\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(2A=31\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(2A=31\left(2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(2A=31\left(2-\frac{1}{13}\right)\)
\(2A=31.\frac{25}{13}\)
\(2A=\frac{775}{13}\)
\(\Rightarrow A=\frac{775}{13}:2\)
\(A=\frac{775}{26}\)
Thay vào (1) ta có:
\(x-\frac{775}{26}=\frac{9}{13}\)
\(\Leftrightarrow x=\frac{9}{13}+\frac{775}{26}\)
\(\Leftrightarrow x=\frac{61}{2}\)
#)Giải :
\(200-18:\left(372:3x-1\right)-28=166\)
\(\Leftrightarrow200-18:\left(372:3x-1\right)=194\)
\(\Leftrightarrow18:\left(372:3x-1\right)=6\)
\(\Leftrightarrow372:3x-1=3\)
\(\Leftrightarrow3x-1=124\)
\(\Leftrightarrow3x=125\)
\(\Leftrightarrow x=\frac{125}{3}\)
200 - 18 : (372 : 3 . x - 1) - 28 = 166
=> 200 - 18 : (372 : 3.x - 1) = 166 + 28
=> 200 - 18 : (372 : 3.x) - 1) = 194
=> 18 : (372 : 3.x - 1) = 200 - 194
=> 18 : (372 : 3.x - 1) = 6
=> 372 : 3.x - 1 = 18 : 6
=> 372 : 3.x - 1 = 3
=> 372 : 3.x = 3 + 1
=> 372 : 3.x = 4
=> 3.x = 372 : 4
=> 3.x = 93
=> x = 93 : 3
=> x = 31
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
a) \(\frac{24}{31}\): (\(\frac{2}{31}\)* 3) = 4
b) (\(\frac{12}{5}\): 2) : ( \(\frac{3}{30}\)*2) = 6
Ta có : 2/5 = 14/35
=> Phải thêm 35 - 31 = 4 ( đơn vị) vào mẫu số và giữ nguyên tử số thì được phân số mới bằng 2/5.
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{101}{102}\)
\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{101}{102}\)
\(1-\frac{1}{x+2}=\frac{101}{102}\)
\(1-\frac{1}{x+2}=1-\frac{1}{102}\)
\(\frac{1}{x+2}=\frac{1}{102}\)
x+2=102
x=102-2
x=100
2/1x3 + 2/3x5 + 2/5x7 + ... + 2/Xx(X+ 2 ) = 101/102
1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .. + 1/x - 1/x + 2 = 101/102
1 - 1/x + 2 = 101/102
1 - 1/x + 2 = 1 - 1/102
1/x + 2 = 1/102
x + 2 =102
x = 102 - 2
x = 100
Chúc bạn học tốt!
a) \(\dfrac{25}{37}\times\dfrac{18}{29}+\dfrac{18}{29}\times\dfrac{12}{37}\)
\(=\dfrac{18}{29}\times\left(\dfrac{25}{37}+\dfrac{12}{37}\right)\)
\(=\dfrac{18}{29}\times\dfrac{37}{37}\)
\(=\dfrac{18}{29}\times1\)
\(=\dfrac{18}{29}\)
b) \(\dfrac{31}{85}\times\dfrac{11}{19}+\dfrac{31}{85}\times\dfrac{12}{19}-\dfrac{42}{19}\times\dfrac{31}{85}\)
\(=\dfrac{31}{85}\times\left(\dfrac{11}{19}+\dfrac{12}{19}-\dfrac{42}{19}\right)\)
\(=\dfrac{31}{85}\times\dfrac{-19}{19}\)
\(=\dfrac{31}{85}\times-1\)
\(=-\dfrac{31}{85}\)
c) \(\dfrac{16}{53}:\dfrac{17}{9}-\dfrac{16}{53}:\dfrac{17}{8}\)
\(=\dfrac{16}{53}:\left(\dfrac{9}{17}-\dfrac{8}{17}\right)\)
\(=\dfrac{16}{53}:\dfrac{1}{17}\)
\(=\dfrac{16}{901}\)
c) \(\dfrac{1}{5}\times\dfrac{12}{31}\times\dfrac{4}{3}+\dfrac{19}{31}\times\dfrac{4}{15}\)
\(=\dfrac{4}{15}\times\dfrac{12}{31}+\dfrac{19}{31}\times\dfrac{4}{15}\)
\(=\dfrac{4}{15}\times\left(\dfrac{12}{31}+\dfrac{19}{31}\right)\)
\(=\dfrac{4}{15}\times\dfrac{31}{31}\)
\(=\dfrac{4}{15}\times1\)
\(=\dfrac{4}{15}\)
a: =18/29*(25/37+12/37)
=18/29
b: =31/85(11/19+12/19-42/19)
=-31/85
c; =16/53(9/17+8/17)=16/53
d: =4/15(12/31+19/31)=4/15
\(X=\frac{908}{75}\)kq đó
Ta có: x-(\(\frac{31}{5}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}\)\(+\frac{31}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)-\(\frac{31}{2}\)x(\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{13}\right)\)=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)\(-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}.\frac{10}{39}\)\(=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{155}{39}=\frac{9}{13}\)
x-\(\frac{434}{195}\)=\(\frac{9}{13}\)
x =\(\frac{9}{13}+\frac{434}{195}=\frac{569}{195}\)
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