Ta có 5=2+3; 9=4+5: 13=6+7: 17=8+9......

Do vậy x=a + ( a+1 ) ( a thuộc N )

Nên 1+5+9+13+16.....+x=1+2+3+4+5+6+7.....+a+(a+1)=501501

Hay (a+1)(a+2)=1003002=1001.1002

==>a=1000

do đó x=1000+(1000+1)=2001

(x+1)x:2=501.1001. =>(x+1)x=1002.1001=>x=1001

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Ta co :

x+1:4+1=501501

x+1:4    = 501501-1

x+1:4    =501500

x+1      = 501500:4

x+1      = 125375

x         = 125375 -1

x         =125374

Vay x=125735 

**** nhe 

\(1+5+9+...+x=501501\)

\(\Leftrightarrow\frac{\left(1+x\right).\left[\left(x-1\right)\div4+1\right]}{2}=501501\)

\(\Leftrightarrow\left(1+x\right).\left[\left(x-1\right)\div4+1\right]=501501.2\)

\(\Leftrightarrow\left(1+x\right).\left[\frac{x-1}{4}+1\right]=1003002\)

\(\Leftrightarrow\frac{x-1}{4}\left(1+x\right)+x+1=1003002\)

\(\Leftrightarrow\frac{x-1}{4}+\frac{x-1}{4}x+x=1003001\)

\(\Leftrightarrow\frac{x-1+x\left(x-1\right)+4x}{4}=1003001\)

Tự giải tiếp

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Hok tốt !!!!

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Nick kia mk đg bj báo cáo nên mk đành pải vào nick này !!!

Nhớ ha !!! Hok tốt !!!!

Giải :

1+5+9+13+17+...+x=50151

ta có 1=0+1

         5=2+3

         9=4+5

         13=6+7

         17=8+9

     .................

=>x= a+(a+1)

nên 1+5+9+...+x=501501

(=)0+1+2+3+4+5+....+a+(a+1)=501501

(=) [(a+1).9a+1+10]:2=501501

(=)(a+1).(a+2)1003002-1001.1002

=>a=1000

=>x=2001

Ta có: 5 = 2+3; 9 = 4+5 ; 13 = 6+7 ; 17 = 8+9;.....

Do vậy x = a + ( a + 1) ( a thuộc N )

Nên 1 + 5 + 9 + 13 + 16 + ....+ x = 1+2+3+4+5+6+7+.....+a+ ( a + 1 ) = 501501

Hay (a + 1)( a + 2) = 1003002 = 1001 . 1002

Suy ra : a = 1000

Do đó : x = 1000 + ( 1000+ 1) = 2001

Vậy, x= 2001

vi sao ma (a+1)(a+2) = 1003002

Ta có: \(\left(x+1\right).x\div2=501501\)

          \(\left(x+1\right).x=501501.2\)

          \(\left(x+1\right).x=10003002\)

          \(\left(x+1\right).x=10003002=\)

Ta có: 5 = 2+3; 9 = 4+5 ; 13 = 6+7 ; 17 = 8+9;.....

Do vậy x = a + ( a + 1)  ( a thuộc N )

Nên 1 + 5 + 9 + 13 + 16 + ....+ x = 1+2+3+4+5+6+7+.....+a+ ( a + 1 ) = 501501

Hay (a + 1)( a + 2)  = 1003002 = 1001 . 1002 

Suy ra : a = 1000

Do đó : x = 1000 + ( 1000+ 1) = 2001

tao ko biet

x=2001

TA có : 5 = 2 + 3 ; 9 = 4 + 5 ; 13 = 6 + 7 ; 17 = 8 + 9 ; ....

Do vậy x = a + ( a + 1 ) a thuộc N 

Nên 1 + 5 + 9 + 13 + 17 + ..+ x  = 1+ 2 + 3 + 4 +5 + 6 + 7 +...+ a +( a+ 1 ) = 501501 

Hay ( a + 1 )(a + 2 ) = 1003002 = 1001.1002 

Do đó x = 1000 ( 1000 + 1 ) = 2001