Tim x
1/2.3 . x +1/3.4 .x + ... + 1/49.50 . x=1
K
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Những câu hỏi liên quan
VK
12 tháng 6 2018
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\frac{49}{50}x=\frac{49}{50}\)
\(x=\frac{\frac{49}{50}}{\frac{49}{50}}\)
\(x=1\)
Vậy \(x=1\)
NA
20 tháng 3 2022
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
VD
20 tháng 3 2022
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
TH
4
8 tháng 5 2017
1.Tính
\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(E=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(E=\frac{1}{1}-\frac{1}{50}\)
\(E=\frac{49}{50}\)
Câu 2 mình không biết, xin lỗi nha
LT
3
5 tháng 5 2019
\(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\frac{2}{3}x-\frac{1}{3}=\frac{1}{2}-\frac{2}{3}\)
\(\frac{2}{3}x-\frac{1}{3}=\frac{-1}{6}\)
\(\frac{2}{3}x=\frac{-1}{6}+\frac{1}{3}\)
\(\frac{2}{3}x=\frac{1}{6}\)
\(x=\frac{1}{6}:\frac{2}{3}\)
\(x=\frac{1}{4}\)
~ Hok tốt ~
KN
5 tháng 5 2019
\(\frac{3}{x+5}=15\%\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{15}{100}\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow x+5=20\)
\(\Leftrightarrow x=20-5\)
\(\Leftrightarrow x=15\)
DL
24 tháng 8 2019
1/1.2+1/3.4+1/5.6+...+1/49.50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=1/1+1/2+1/3+1/4+...+1/49+1/50-2(1/2+1/4+1/6+...+1/50)
=1/1+1/2+1/3+1/4+...+1/49+1/50-(1/1+1/2+1/3+1/4+...+1/25)
=1/26+1/27+...+1/50=1/26+1/27+...+1/50(đpcm)
b. 1/1-1/2+1/3-1/4+...+1/99-1/100=99/100
7/12=175/300; 5/6=10/12=250/300; 99/100=297/300
(hình như khúc này đề bài sai hả bạn) bạn tự tính ra nhé
bài 2: a.x+1/10+x/12+x/14+...x+1/20
(x+x+x...+x)+(1/10+1/12+...+1/20)
ko có kết quả sao tìm x được bạn:[
b.x+1/2000+x+2/1999=x+3/1998+x+4/1997
x+1/2000+x+2/1999=x+3/1998+x+4/1997
(x+1/2000+1)+(x+2/1999+1)=(x+3/1998+1)+(x+4/1997+1)
x+2002/2000+x+2002/1999=x+2002/1998+x+2002/1997
x+2002(1/2000+1/1999)=(x+2002)(1/1998+1/1997)
=>(1/2000+1/1999)=(1/1998+1/1997)
x+2002(1/2000+1/1999)-(x+2002)(1/1998+1/1997)=0
(x+2002)(1/2000+1/1999-1/1998-1/1997)=0
(x+2002).0=0
(x+2002)=0
x =0-2002=-2002
Chúc bạn học tốt.
MV
21 tháng 8 2017
\(x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{49\cdot50}=\dfrac{25}{13}\\ x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)=\dfrac{25}{13}\\ x-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{25}{13}\\ x-\left(1-\dfrac{1}{50}\right)=\dfrac{25}{13}\\ x-\dfrac{49}{50}=\dfrac{25}{13}\\ x=\dfrac{25}{13}+\dfrac{49}{50}\\ x=\dfrac{1887}{650}\)
27 tháng 6 2016
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{13}{25}x=1\)
\(x=1:\frac{13}{25}=\frac{25}{13}\)
27 tháng 6 2016
( 1/2x3 +1/3x4 + ... + 1/49x50 ) x X = 1
( 3-2/2x3 + 4-3/3x4 + ... + 50-49/49x50 ) x X = 1
( 1/2 -1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50 ) x X = 1
( 1/2 - 1/50 ) x X = 1
12/25 x X = 1
X = 1 : 12/25
X = 25/12
PV
7
17 tháng 5 2017
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\dfrac{12}{25}x=1\)
\(\Rightarrow x=\dfrac{25}{12}\)
Vậy \(x=\dfrac{25}{12}\)
AT
17 tháng 5 2017
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right).x=1\)
Ta có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(=\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}\)
\(\Rightarrow\dfrac{12}{25}.x=1\Rightarrow x=1:\dfrac{12}{25}=\dfrac{25}{12}=2\dfrac{1}{12}\)
Vậy \(x=\dfrac{25}{12}\) hay \(x=2\dfrac{1}{12}\)
=>(1/1.2+ 1/2.3+1/3.4+1/4.5+...+1/49.50 ) .x =1
=>(1/1 -1/2 +1/2 -1/3 +1/3-1/4+.......+1/49 -1/50).x =1
=>( 1 -1/50 ) .x= 49/50 .x = 1
=> x=1:49/50=50/49