Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
|x+2| + x = 4
=> | x + 2| = 4 - x
Ta có các trường hợp :
TH1 : x + 2 = 4 - x
=> 2x = 4 - 2
=> x = 1
Th2 : x + 2 = x - 4
=> x - x = -4 - 2 (vô lí)
Vậy ...
Qúa dễ lun chứ!
1 + 2 + 1 = 4
Còn đòi Toán 6 cái gì?
![](https://rs.olm.vn/images/avt/0.png?1311)
1,
x10 = x
=> x10 - x = 0
=> x(x9 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
KL: x thuộc {1; 0}
2,
\(S=2+2^2+2^3+...+2^{2016}\)
=> \(2S=2^2+2^3+2^4+...+2^{2017}\)
=> \(2S-S=\left(2^2+2^3+2^4+...+2^{2017}\right)-\left(2+2^2+2^3+...+2^{2016}\right)\)
=> \(S=2^{2017}-2\)
Bài 1:
x10 = x => x= { -1;1}
Bài 2:
\(S=2+2^2+2^3+...+2^{2016}\)
\(2S=2^2+2^3+2^4+2^{2017}\)
\(2S-S=2^{2017}-2\)
Vậy \(S=2^{2017}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(2^0+2^1+2^2\right)+...+\left(2^{2001}+2^{2002}+2^{2003}\right)+2^{2004}+2^{2005}=7+2.7+...+2^{2001}.7+2^{2004}+2^{2005}\)Ta đi tìm số dư khi chia \(2^{2004}+2^{2005}=8^{668}+2.8^{668}\equiv1+2.1\equiv3\left(mod7\right)\)
Vậy B chia 7 dư 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Rightarrow A=\frac{1}{4}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Rightarrow A=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{24360}\right)\)
\(\Rightarrow A=\frac{1353}{32480}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}x-\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{5}{12}\)
\(\Rightarrow\frac{2}{3}x-\frac{3}{2}x+\frac{3}{4}=\frac{5}{12}\)
\(\Rightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}-\frac{3}{4}\)
\(\Rightarrow-\frac{5}{6}x=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right):\left(-\frac{5}{6}\right)\)
\(\Rightarrow x=\frac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : S = 1 - 2 + 22 + 23 + ...... + 21000
=> S = -1 + (2 + 22 + 23 + ...... + 21000)
Đặt G = 2 + 22 + 23 + ...... + 21000
=> 2G = 22 + 23 + ...... + 21001
=> 2G - G = 21001 - 2
=> G = 21001 - 2
Nên : S = -1 + 21001 - 2
=> S = 21001 - 3
Có gì đó sai sai 1 - 2 + 2^2 + 2^3 + .... + 2^998 - 2^999 + 2^1000 là sao
1 + 2 + 2^2 + 2^3 + .... + 2^998 + 2^999 + 2^1000 hay
1 - 2 + 2^2 - 2^3 + .... + 2^998 - 2^999 + 2^1000