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#)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
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 =\(\frac{9}{13}+\frac{434}{195}=\frac{569}{195}\)
nhé
\(=\frac{\frac{28}{31}.\frac{31}{7}}{\frac{8}{9}.\frac{9}{4}}=\frac{4}{2}=2\)
Ta có \(\frac{x}{13}=\frac{35}{31}\)
<=>31x=13.35
<=>31x=455
<=>x=\(\frac{455}{31}\)
x.31 = 13.35
31x = 455
x = 455:31
Vậy x = 445/31
NHớ k cho mình nhé!
bạn Kudo Shinichi sai rồi: muốn tìm số chia mà thương chia cho số bị chia à
a) 7,2 * x = 6,49
x = 6,49 : 7,2
x = 649/720
b) \(\frac{15}{77}:x=\frac{3}{11}\)
\(x=\frac{3}{11}:\frac{15}{77}=\frac{7}{5}\)
c) 2,4 * x + 1,1 *x = 0,7
x * ( 2,4 + 1,1 ) = 0,7
x * 3,5 = 0,7
=> x = 0,7 : 3,5 = 0,2
phần d sai đề đó bạn, 25/31 chia rồi lại công là sai, sửa đề đi nha.
Ai thấy đúng thì ủng hộ nha !!!
<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15
[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15
[2.(1/3-1/15)]+x=17/15
(2.4/15)+x=17/15
6/15+x=17/15
x=17/15-6/15
x=11/15
\(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}=\frac{31}{1.3}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}+\frac{31}{11.13}\\ \)
\(=\frac{31}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{13}\right)=\frac{31}{2}.\frac{12}{13}=\frac{31.6}{13}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\Leftrightarrow x=\frac{195}{13}=15\)
\(x-\left(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\)\(\frac{9}{13}\)(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:
\(x-\frac{775}{26}=\frac{9}{13}\)
\(\Leftrightarrow x=\frac{9}{13}+\frac{775}{26}\)
\(\Leftrightarrow x=\frac{61}{2}\)