Coi x - 10 là y.
=> x - 9 là y + 1
........... x - 1 là y + 9.
Ta có : (x-1)+(x-2)+...+(x-10) = 165
<=> (y+9)+(y+8)+...+y = 165.
=> 10y + 45 = 165.
=> 10y = 120.
=> y = 12.
Tương đương : x - 1 = 12 + 9
=> x = 21 + 1 = 22.
Vậy x là 22.
Coi \(x-10=y\)
\(\Rightarrow x-9=y+1\)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-10\right)=165\)
Ta có hai số hạng liền trước số 10
10 - 1 = 9 ; 9 - 1 = 8
\(\Leftrightarrow\left(y+9\right)+\left(y+8\right)+...+y=165\)
\(=10y+45=165\)
\(\Rightarrow10y=165-45=120\)
\(\Rightarrow y=120:10=12\)
Ta có x - 1
\(\Rightarrow x=21+1=22\)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-10\right)=165\)
\(\Leftrightarrow\left(x+x+...+x\right)-\left(1+2+...+10\right)=165\)
\(\Leftrightarrow10x-55=165\)
\(\Leftrightarrow10x=220\Rightarrow x=22\)
Ta có :
\(\left(x-1\right)+\left(x-2\right)+..........+\left(x-10\right)=165\)
\(\Leftrightarrow\left(x+x+.....+x\right)-\left(1+2+....+10\right)=165\)
\(\Leftrightarrow10x-55=165\)
\(\Leftrightarrow10x=110\)
\(\Leftrightarrow x=10\)
Vậy ..................
C1: \(\left(x-1\right)+\left(x-20\right)+...+\left(x-10\right)=165\)
\(\left(x+x+...+x\right)-\left(1+2+...+10\right)=165\)
\(\left(x.10\right)+\left(1+10\right).10:2=165\)
\(\left(x.10\right)+55=165\)
\(x.10=165-55\)
\(x.10=110\)
\(x=110:10\)
\(x=11\)
Vậy \(x=11\)
C2: \(\left(x-1\right)+\left(x-20\right)+...+\left(x-10\right)=165\)
\(\Rightarrow\) \(10x-\left(1+2+...+10\right)=165\)
\(\Rightarrow\) \(10x+55=165\)
\(\Rightarrow\) \(10x=165-55\)
\(\Rightarrow\) \(10x=165-55\)
\(\Rightarrow\) \(10x=110\)
\(\Rightarrow\) \(x=110:10\)
\(\Rightarrow\) \(x=11\)
Vậy \(x=11\)