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Sửa đề: \(7^{2000}+7^{2001}\)
\(7^{2000}+7^{2001}=7^{2000}\cdot1+7^{2000}\cdot7\)
\(=7^{2000}\left(1+7\right)\)
\(=7^{2000}\cdot8⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
10 + 10 +......+ 10. 0 ( n số 10 )
= 10 + 10 +...+ 10. 0 ( n - 1 số 10 )
= ( n - 1 ) . 10 nha
k nha mình đang âm điểm!
Gợi ý mí bn nè :
10 + 10 + ... + 10 . 0 ( n số 10 )
= 10 + 10 + ... + 10 + 0 ( n - 1 số 10 )
= ( n - 1 ) . 10 nghe
k mk mk k lại cho , mk vẫn chưa có đ hỏi đáp
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a, 7\(x\).(2\(x\) + 10) = 0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\){-5; 0}
b, - 9\(x\) : (2\(x\) - 10) = 0
- 9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
d, (\(x\) + 2023).(\(x\) - 2024) = 0
\(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-2023; 2024}
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
![](https://rs.olm.vn/images/avt/0.png?1311)