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8 tháng 8 2017

ta có:

\(1+2+3+....+x=210\)

\(\Rightarrow\frac{\left(x+1\right).\left(x-1+1\right)}{2}=210\)

\(\Rightarrow\left(x+1\right).x=420\)

ta có: \(20.21=420\Rightarrow x=20\)

8 tháng 8 2017

tích giúp mình nhé

20 tháng 7 2017

ta có : x+3/5+x+4/5=x+5/3+x+6/2

   => (x+x)+(3/5+4/4)=(x+x)+(5/3+6/2)

   => 2x+8/5=2x+14/3  ( vô lí ) 

20 tháng 7 2017

Ta có: \(x+\frac{3}{5}+x+\frac{4}{4}=x+\frac{5}{3}+x+\frac{6}{2}\)

=> \(x+\frac{3}{5}+x+1=x+\frac{5}{3}+x+3\)

=> \(x+1+x+3=x+\frac{5}{3}-x-\frac{3}{5}\)

=> \(2x+4=\frac{25-9}{15}\)

=> \(2x+4=\frac{16}{15}\)

=> \(2x+4=1+\frac{1}{15}\)

=> \(2x+4-1=\frac{1}{15}\)

=> \(2x+3=\frac{1}{15}\)

=> \(2x=\frac{1}{15}-3\)

=> \(2x=\frac{1-45}{15}\)

=> \(2x=-\frac{44}{15}\)

=> \(x=\frac{\left(-\frac{44}{15}\right)}{2}\)

=> \(x=-\frac{44}{15}.\frac{1}{2}\)

=> \(x=-\frac{22}{15}\)

31 tháng 5 2020

(x+1)+(x+2)+(x+3)+...+(x+100)=2020

x+1+x+2+x+3+...+x+100=2020

(x+x+x+...+x)+(1+2+3+...+100)=2020

100x+5050=2020

100x=2020-5050

100x=-3030

      x=-30,3

Bài 1: 

a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)

b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)

27 tháng 4 2017

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2009}{2011}\)

Đặt tổng vế trái là A

Ta có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\)

\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)\div2}\right)\)

\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)

\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)

\(A=\left(\frac{1}{2}+\frac{1}{x+1}\right):\frac{1}{2}\)

\(A=1+\frac{1}{\left(x+1\right)\div2}\)

\(\Rightarrow1+\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}-1=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)

\(\Rightarrow-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)

\(\Rightarrow-\left(x+1\right)=2011\)

\(\Rightarrow x+1=-2011\) 

\(\Rightarrow x=-2011-1=-2012\)

a: 3x-2=2x-3

=>x=-1

b: 2x+3=5x+9

=>-3x=6

=>x=-2

c: 5-2x=7

=>2x=-2

=>x=-2

d: 10x+3-5x=4x+12

=>5x+3=4x+12

=>x=9

e: 11x+42-2x=100-9x-22

=>9x+42=78-9x

=>18x=36

=>x=2

f: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5