2x-1/2-1/6-1/12-...-1/49.50=7-1/50+xphần cuối là (1/50)+ x riêng nha...
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2x-1/2-1/6-1/12-...-1/49.50=7-1/50+x
phần cuối là (1/50)+ x riêng nha mng
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TT
1
Những câu hỏi liên quan
26 tháng 5 2022
\(2x-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-...-\dfrac{1}{49\cdot50}=7-\dfrac{1}{50}+x\)
\(\Leftrightarrow2x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=x+\dfrac{349}{50}\)
\(\Leftrightarrow2x-\dfrac{49}{50}-x-\dfrac{349}{50}=0\)
=>x=398/50=199/25
NL
1
HT
14 tháng 7 2015
2x - \(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-....-\frac{1}{49.50}\)= 7-\(\frac{1}{50}\)+x
2x - x - \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\right)\)= 7 - \(\frac{1}{50}\)
x - \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)= \(\frac{349}{50}\)
x - \(\left(1-\frac{1}{50}\right)\)=\(\frac{349}{50}\)
x - \(\frac{49}{50}\)=\(\frac{349}{50}\)
x = \(\frac{349}{50}+\frac{49}{50}\)
x = \(\frac{199}{25}\)
4 tháng 9 2017
2x - \(\dfrac{1}{2}-\dfrac{1}{6}-...-\dfrac{1}{49.50}\)= 6-\(\dfrac{1}{50}\) + x
<=> x - ( \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)) = \(\dfrac{299}{50}\)
<=> x - \(\left(1-\dfrac{1}{50}\right)\) = \(\dfrac{299}{50}\)
<=> x - \(\dfrac{49}{50}\) = \(\dfrac{299}{50}\)
<=> x = \(\dfrac{174}{25}\)
4 tháng 9 2017
\(2x-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-....-\dfrac{1}{49.50}=7-\dfrac{1}{50}+x\)
\(\Rightarrow2x-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{49.50}\right)=7-\dfrac{1}{50}+x\)
\(\Rightarrow2x-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=7-\dfrac{1}{50}+x\)
\(\Rightarrow2x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=7-\dfrac{1}{50}+x\)\(\Rightarrow2x-1+\dfrac{1}{50}=7-\dfrac{1}{50}+x\)
\(\Rightarrow2x=7-\dfrac{1}{50}+x-\dfrac{1}{50}+1\)
\(\Rightarrow2x=\dfrac{199}{25}+x\)
\(\Rightarrow x=\dfrac{199}{25}\)
18 tháng 6 2017
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-....-\frac{1}{49.50}=7+\frac{1}{50}+x\)
\(2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{49.50}\right)=7+\frac{1}{50}+x\)
\(2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\right)=7+\frac{1}{50}+x\)
\(2x-\left(\frac{1}{1}-\frac{1}{50}\right)=7+\frac{1}{50}+x\)
\(2x-1+\frac{1}{50}=7+\frac{1}{50}+x\)
=> 2x - 1 = 7 + x
=> 2x - x = 7 + 1
=> x = 8
NT
0
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.
Ta có:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
Phương trình ban đầu tương đương với:
\(2x-\frac{49}{50}=7-\frac{1}{50}+x\)
\(\Leftrightarrow x=7-\frac{1}{50}+\frac{49}{50}=\frac{198}{25}\)