Tìm x thuộc N
(x-10)3= (x-10)100
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a) \(???\)
b) \(123x+877x=2000\)
\(1000x=2000\)
\(x=2000:1000\)
\(x=2\)
c) \(2x.\left(x-10\right)=0\)
=> \(x-10=0\)
\(x=10\)
d)\(6.\left(x+2\right)-\left(4x+10\right)=100\)
\(6.x+12-4x+10=100\)
\(2x+2=100\)
\(2x=98\)
\(x=98:2\)
\(x=49\)
e) \(x.\left(x+1\right)=2+4+6+8+...+2500\)
\(x.\left(x+1\right)=1563750\)
mà ta thấy : \(1250.1251=1563750\)
=> \(x=1250\)
g)\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(x.100+5050=5750\)
\(x.100=5750-5050\)
\(x.100=700\)
\(x=7\)
493 ⋮ x
⇒ x ∈ Ư(493) = {1; 17; 29; 493}
Mà 10 < x < 100
⇒ x ∈ {17; 29}
10+100+2010+x
= 2120+x
2120 chia hết cho 2
=> x phải là số lẻ
\(\left(x-10\right)^5=\left(x-10\right)^3\) (Sửa dấu \(-\rightarrow=\))
\(\Rightarrow\left(x-10\right)^5-\left(x-10\right)^3=0\)
\(\Rightarrow\left(x-10\right)^3\left[\left(x-10\right)^2-1\right]=0\)
\(\Rightarrow\left(x-10\right)^3\left[\left(x-10+1\right)\left(x-10-1\right)\right]=0\)
\(\Rightarrow\left(x-10\right)^3\left(x-9\right)\left(x-11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-10=0\\x-9=0\\x-11=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\\x=11\end{matrix}\right.\)
a, A={11,12,13,14,15}
b, B={10,11,12,13,14,15,16,17,18,19,20}
c, C={6,7,8,9,10}
d,D={11,12,13,...,95,96,97,98,99,100}
e, E={2983,2984,2985,2986}
f, F={0,1,2,3,4,5,6,7,8,9}
g, G={0,1,2,3,4}
h, H={0,1,2,3,4,5,6,...,98,99,100}
493 chia hết cho x => x \(\in\)Ư(493) = {1;17;29;493}
Mà 10 < x < 100 => x \(\in\){17;29}
Vậy x \(\in\){17;29}
\(493⋮x\)\(\Rightarrow x\inƯ\left(493\right)=\left\{1;17;29;493\right\}\)
mà \(10< x< 100\)\(\Rightarrow x\in\left\{1;17;29\right\}\)
Vậy \(x\in\left\{1;17;29\right\}\)
\(\Leftrightarrow\left(x-3\right)^8-\left(x-3\right)^{10}=0\)
\(\Leftrightarrow\left(x-3\right)^8\left[1-\left(x-3\right)^2\right]=0\)
\(\Leftrightarrow\left(x-3\right)^8=0hay1-\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0hay\left(x-3\right)^2=1\)
\(\Leftrightarrow x=3hay\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\)
\(\Leftrightarrow x=3hay\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
(x-10)^3 = (x-10)^100
=> (x-10)^3 - (x-10)^100 = 0
(x-10)^3. ( 1 - (x-10)^97) = 0
=> (x-10)^3 = 0 => x- 10 = 0 => x = 10
1- (x-10)^97 = 0 => (x-10)^97 = 1 => x - 10 = 1 => x = 11
KL: ...