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1 + 1 + 1 = 13
1 + 1 + 2 = 24
1 + 1 + 3 = 35
1 + 1 + 6 = 68
k mk nhé, mk đang âm điểm ghê lắm
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\(\dfrac{193}{140}\)
1, 2x - 35 = 15
2x = 15 + 35
2x = 50
x = 50 : 2
x = 25.
2, 3x + 18 = 12
3x = 12 - 18
3x = -6
x = -6 : 3
x = -2.
3, / x - 1 / = 0
=> x \(\in\varnothing\).
4, -13 /x/ = - 26
/x/ = -26 : -13
=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy x \(\in\){ 2 ; -2}.
5,4 - ( 27 - 3 ) = x - ( 13 - 4 )
4 - 24 = x - 9
-20 = x - 9
-x = 9 + 20
-x = 29
x = -29.
6, 47 - ( x + 15 ) = 21
47 - x - 15 = 21
-x - 15 = 21 - 47
-x - 15 = -26
-x = -26 + 15
-x = - 11
x = 11.
7, -5 -( 24 - x) = - 11
-5 - 24 + x = -11
-24 + x = -11 + 5
-24 + x = -6
x = -6 + 24
x = 18.
8, 6 - /x/ = 2
/x/ = 6 - 2
\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy x \(\in\left\{3;-3\right\}.\)
9, 6 + /x/ = 2
/x/ = 2 - 6
=> x = -4.
2x - 35 = 15
=> 2x = 15 + 35
=> x = 50 : 2
=> x = 25
3x + 18 = 12
=> 3x = 12 - 18
=> x = ( -6 ) : 3
=> x = -2
| x - 1 | = 0
=> x - 1 = 0
=> x = 0 + 1
=> x = 1
-13 * | x | = -26
=> | x | = -26 : ( -13 )
=> | x | = 2
96 : ( 3,75 : x ) = 94,5
3,75 : x = 96 : 94,5
3,75 : x = \(\frac{96}{94,5}\)
x = 3,75 : \(\frac{96}{94,5}\)
x = \(\frac{354,375}{96}\)
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{x}\left(3+x\right)=\frac{6}{13}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{3+x}{x}=\frac{6}{13}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+1+\frac{3}{x}=\frac{6}{13}\)
\(\Rightarrow\frac{3}{x}+\frac{8}{5}=\frac{6}{13}\)
\(\Rightarrow\frac{3}{x}=-\frac{74}{65}\)
\(\Rightarrow-74x=195\)
\(\Rightarrow x=\frac{-195}{74}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Mình chỉnh lại đề B nha:
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
Ta có :
a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\)\(\frac{1}{2}.\frac{4}{15}\)
\(=\)\(\frac{2}{15}\)
Ta có :
\(c)\)\(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+\frac{37}{1000}+...+\frac{229}{1000}\)
\(=\)\(\frac{1+13+25+37+...+229}{1000}\)
Xét tổng \(1+13+25+37+...+229\):
Số số hạng : \(\left(229-1\right):12+1=20\) ( số hạng )
Tổng : \(\frac{\left(229+1\right).20}{2}=2300\)
Do đó :
\(\frac{1+13+25+37+...+229}{1000}=\frac{2300}{1000}=\frac{23}{10}\)
1+1+1=13
1+1+2=24
1+1+3=35
...
1+1+6 = 48