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b, x+x-1+x-2+...+x-50=255
=>x+x+x+...+x-1-2-3-4-...-50=255
=>x.51-(1+2+3+...+50)=255
=>x.51-[(50+1).(50-1+1):2]=255
=>x.51-[51.50:2]=255
=>x.51-1275=255
=>x.51=255+1275=1530
=>x=1530:51=30
Vậy x = 30
a, x + 1/2x - 25%x = 10
=> x + 1/2x - 1/4x = 10 => x.( 1/2 - 1/4 ) =10 => x.1/4 = 10 =>x = 10 : 1/4 = 40 Vậy x = 4051x - ( 1+2+3+4+....+50) = 255
51x - 1275 = 255
51x = 255 + 1275
51x = 1530
x= 1530 : 51
x =30
x+x-1+x-2+x-3+x-4+...+x-50 = 255
- 51x - (50 + 49 + 48 + ....+ 1) = 255
- 51x - 1275 = 255
- 51x 255 + 1275
- 51x = 1530
- x = 1530 : 51
- x = 30
\(x-1+x-2+x-3+...................+x-50=225\)
\(\Rightarrow\left(x+x+x+...............+x\right)-\left(1+2+3+..............+50\right)=225\)
\(\Rightarrow50x-1275=225\)
\(\Rightarrow50x=1500\)
\(\Rightarrow x=30\)
\(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+1=16\\x+1=-16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-17\end{cases}}\)
x+x-1+x-2+x-3+....+x-50 = 255
x+x+x+...+x - (1+2+3+...+50) = 255
51x - 1275 = 255
=> 51x = 1530
=> x = 30
a, \(6⋮x-1\)hay \(x-1\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
x - 1 | 1 | 2 | 3 | 6 |
x | 2 | 3 | 4 | 7 |
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ..... + ( x + 10 ) = 255
=> x + 1 + x + 2 + x + 3 + .................. + x + 10 = 255
=> 10x + ( 1 + ........... + 10 ) = 255
10x + 55 = 255
10x = 255 - 55 = 200
x = 200 : 10 = 20
Chắc z
<=>10x+(1+2+3+4+5+6+7+8+9+10)=255
<=> 10x+55=255
<=>10x = 255-55
<=> 10x=200
=> X=200:10=20