Đáp án:
4.
a, Ta có :
$2x + 4x + 6x + .... + 98x = 4900$
$ <=> x.(2 + 4 + 6 + .... + 98) = 4900$
$ <=> x . 2450 = 4900$
$ <=> x = 2$
b, Ta có :
$7x + 9x + 11x + .... + 121x = 11 136$
$ <=> x . ( 7 + 9 + 11 + .... + 121) = 11136$
$ <=> x . 3712 = 11136$
$ <=> x = 3$
c, Ta có :
$\dfrac{1}{1.2}$+$\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + .... + $\dfrac{1}{x.(x + 1)}$ =$\dfrac{199}{200}$
<=> 1 - $\dfrac{1}{2}$ +$\dfrac{1}{2}$ - $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{4}$ + ..... + $\dfrac{1}{n}$ - $\dfrac{1}{n+1}$ = $\dfrac{199}{200}$
<=> 1 - $\dfrac{1}{n + 1}$ = $\dfrac{199}{200}$
<=> $\dfrac{1}{n + 1}$ = 1 - $\dfrac{199}{200}$
<=> $\dfrac{1}{n + 1}$ = $\dfrac{1}{200}$
$<=> n + 1 = 200$
$ <=> n = 199$
5.
a, Ta có :
$\dfrac{x + 2}{98}$ + $\dfrac{x + 3}{97}$ + $\dfrac{x + 4}{96}$ + $\dfrac{x + 5}{95}$ =- 4
<=> $\dfrac{x + 2}{98}$ + $\dfrac{x + 3}{97}$ + $\dfrac{x + 4}{96}$ + $\dfrac{x + 5}{95}$ + 4 = 0
<=> $\dfrac{x + 2}{98}$ + $\dfrac{x + 3}{97}$ + $\dfrac{x + 4}{96}$ + $\dfrac{x + 5}{95}$ + ( 1 + 1 + 1 + 1) = 0
<=> ($\dfrac{x + 2}{98}$ + 1) + ($\dfrac{x + 3}{97}$ + 1) + ($\dfrac{x + 4}{96}$ + 1) + ($\dfrac{x + 5}{95}$ + 1) = 0
<=> $\dfrac{x + 100}{98}$ + $\dfrac{x + 100}{97}$ + $\dfrac{x + 100}{96}$ + $\dfrac{x + 100}{95}$ = 0
<=> ( x + 100).($\dfrac{1}{98}$ + $\dfrac{1}{97}$ + $\dfrac{1}{96}$ + $\dfrac{1}{95}$) = 0
$<=> x + 100 = 0 $
$<=> x = - 100$
b, Ta có :
$\dfrac{x -3 }{203}$ + $\dfrac{x - 4 }{204}$ + $\dfrac{x - 5 }{205}$ = $\dfrac{x + 101}{99}$ + $\dfrac{x + 102}{98}$ + $\dfrac{x + 103}{97}$
<=> $\dfrac{x -3 }{203}$ + $\dfrac{x - 4 }{204}$ + $\dfrac{x - 5 }{205}$ + 4 = $\dfrac{x + 101}{99}$ + $\dfrac{x + 102}{98}$ + $\dfrac{x + 103}{97}$ + 4
<=> $\dfrac{x -3 }{203}$ + $\dfrac{x - 4 }{204}$ + $\dfrac{x - 5 }{205}$ + ( 1 + 1 + 1 + 1)= $\dfrac{x + 101}{99}$ + $\dfrac{x + 102}{98}$ + $\dfrac{x + 103}{97}$ + ( 1 + 1 + 1 + 1)
<=> ($\dfrac{x -3 }{203}$ + 1)+ ($\dfrac{x - 4 }{204}$ + 1)+ ($\dfrac{x - 5 }{205}$ + 1) = ($\dfrac{x + 101}{99}$ + 1)+ ($\dfrac{x + 102}{98}$ + 1)+ ($\dfrac{x + 103}{97}$ + 1)
<=> $\dfrac{x + 200}{203}$ + $\dfrac{x + 200 }{204}$ + $\dfrac{x + 200 }{205}$ = $\dfrac{x + 200}{99}$ + $\dfrac{x + 200}{98}$ + $\dfrac{x + 200}{97}$
<=> ( x + 200)($\dfrac{1}{203}$ + $\dfrac{1}{204}$ + $\dfrac{1 }{205}$) = ( x + 200)( $\dfrac{1}{99}$ + $\dfrac{1}{98}$ + $\dfrac{1}{97}$)
$<=> x + 200 = 0 $
$<=> x = -200$
c, Ta có
$\dfrac{x - 3 }{17}$ + $\dfrac{x - 4}{16}$ + $\dfrac{x - 5}{15}$ + $\dfrac{x - 6}{14}$ + $\dfrac{x - 7}{13}$ =5
<=> $\dfrac{x - 3 }{17}$ + $\dfrac{x - 4}{16}$ + $\dfrac{x - 5}{15}$ + $\dfrac{x - 6}{14}$ + $\dfrac{x - 7}{13}$ - 5 = 0
<=> $\dfrac{x - 3 }{17}$ + $\dfrac{x - 4}{16}$ + $\dfrac{x - 5}{15}$ + $\dfrac{x - 6}{14}$ + $\dfrac{x - 7}{13}$ - ( 1 + 1 + 1 + 1 + 1) = 0
<=>($\dfrac{x - 3 }{17}$ - 1) +( $\dfrac{x - 4}{16}$ - 1) + ($\dfrac{x - 5}{15}$ - 1)+ ($\dfrac{x - 6}{14}$ - 1)+ ($\dfrac{x - 7}{13}$ - 1)=0
<=> $\dfrac{x - 20 }{17}$ + $\dfrac{x - 20}{16}$ + $\dfrac{x - 20}{15}$ + $\dfrac{x - 20}{14}$ + $\dfrac{x - 20}{13}$ =0
<=> ( x - 20).($\dfrac{1}{17}$ + $\dfrac{1}{16}$ + $\dfrac{1}{15}$ + $\dfrac{1}{14}$ + $\dfrac{1}{13}$ ) = 0
$<=> x - 20 = 0 $
$<=> x = 20$
