1. Ta có : 1 + 2 + ....+ x = 210 .
Ta có : \(\left(x-1\right):1+1=x-1+1=x\)
=> Lượng số hạng là x .
Ta có : Tổng \(=\frac{x\left(x+1\right)}{2}=210\)
=> \(x\left(x+1\right)=420\)
=> \(x^2+x-420=0\)
=> \(x^2-20x+21x-420=0\)
=> \(x\left(x-20\right)+21\left(x-20\right)=0\)
=> \(\left(x+21\right)\left(x-20\right)=0\)
=> \(\left[{}\begin{matrix}x+21=0\\x-20=0\end{matrix}\right.\)
Mà theo quy luật dãy số : \(0< x< 210\)
=> \(x-20=0\)
=> \(x=20\)
2, Ta có : \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
=> \(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
=> \(\left(2x-15\right)^3\left(\left(2x-15\right)^2-1\right)=0\)
=> \(\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\end{matrix}\right.\)
Vậy ...
3, Ta có : \(x^{10}=x\)
=> \(x^{10}-x=0\)
=> \(x\left(x^9-1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy ...
=> \(\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)
=> \(\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
=> \(\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy ...
