\(x+2x+3x+...+100x=-213\)
\(\Leftrightarrow x.\left(1+2+3+..+100\right)=-213\)
\(\Leftrightarrow x.[\frac{\left(100+1\right).100}{2}]=-213\)( áp dụng dãy số cách đều )
\(\Leftrightarrow x.5050=-213\)
\(\Leftrightarrow x=\frac{-213}{5050}\)
x + 2x + 3x + ... + 100x = -213
100x + 5050 = -213
100x = -213 - 5050 = -5263
x = -5263 : 100
x = -52,63
\(x+2x+3x+...+100x=-213\)
\(\Rightarrow x\times\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x\times5050=-213\)
\(\Rightarrow x=\frac{-213}{5050}\)
Vậy ...
\(x+2x+3x+...+100x=-213\)
\(\left(x+x+x+...+x\right)\left(1+2+3+...+100\right)=-213\)
\(100x\cdot100=-213\)
\(100x=\frac{-213}{100}\)
\(x=-213\)
