tim so tu nhien ,biet:\(x.\left[x+1\right]=2+4+6+8+...+2500\)
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n.(n+1) = 2 + 4 + 6 + 8 + ...+ 2500 [ có (2500 - 2) : 2 + 1 = 1250 ( số hạng) ]
n.(n+1) = ( 2 + 2500) x 1250 : 2
n.(n+1) = 2502 x 1250 : 2
n.(n+1) = 1250 x 1251
=> n = 1250
Vậy n = 1250
a)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2010\right)=2029099\\ 2011.x+\left(1+2+3+...+2010\right)=2029099\\ 2011.x+2021055=2029099\\ 2011.x=2029099-2021055\\ 2011.x=8044\\ x=8044:2011\\ x=4\)
b)
\(2+4+6+...+2x=210\\ 2.\left(1+2+3+...+x\right)=210\\ 1+2+3+...+x=210:2\\ 1+2+3+...+x=105\\ \dfrac{x.\left(x+1\right)}{2}=105\\ x.\left(x+1\right)=105.2\\ x\left(x+1\right)=210\\ x.\left(x+1\right)=14.15\\\Rightarrow x=14\)
a) x+(x+1)+(x+2)+...+(x+2010)=2029099
x+x+1+x+2+...+x+2010=2029099
2011x+[(2010+1).2010:2]=2029099
2011x+2021055=2029099
2011x=2029099-2021055
2011x=8044
x=8044:2011
x=4
Vậy x=4.
b) 2+4+6+8+...+2x=210
(2x+2)*14:2=210
(2x+2)*7=210
2x+2=210:7
2x+2=30
2x=30-2
2x=28
x=28:2
x=14
Vậy x=14.
con nua nhe
\(\left(=\right)1-x=0hoacx-2=0\)
\(\left(1\right)1-x=0\)
\(\left(=\right)x=1\)
\(\left(2\right)x-2=0\)
\(\left(=\right)x=2\)
vậy x=1;x=2
(x-2)^6-(x-2)^8=0
(x-2)^6[1-(x-2)^2]=0
+) (x-2)^6=0
=>x-2=0
x=2
+) 1-(x-2)^2=0
(x-2)^2=1
+) x-2=1 +) x-2=-1
=>x=3 x=1
(x - 2)6 = (x - 2)8
<=> (x - 2)8 - (x - 2)6 = 0
<=> (x - 2)6(x - 2)2 - (x - 2)6 = 0
<=> (x - 2)6[(x - 2)2 - 1] = 0
<=> \(\orbr{\orbr{\begin{cases}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{cases}}}\)
<=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy ...
Ta có VP(vế phải ) có : (2500-2):2+1=1250(số hạng )
VP =(2500+2)*1250:2=1563750
Theo bài ra ta có :x*[x+1] = 1563750=1250*1251
Vậy x = 1250
\(x\left(x+1\right)=2+4+6+8+..+2500\)
\(\Leftrightarrow x\left(x+1\right)=\frac{2502.1250}{2}=\frac{1251.2.1250}{2}=1250.1251\)
\(\Leftrightarrow x=1250\)