![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x+4}{x+3}< 1\)
\(\Leftrightarrow\frac{x+4}{x+3}-1< 0\)
\(\Leftrightarrow\frac{x+4-x-3}{x+3}< 0\)
\(\Leftrightarrow\frac{1}{x+3}< 0\)
\(\Leftrightarrow x+3< 0\)
\(\Leftrightarrow x< -3\)
Vậy \(x< -3\)
b) \(\frac{x+3}{x+4}>1\)
\(\Leftrightarrow\frac{x+3}{x+4}-1>0\)
\(\Leftrightarrow\frac{x+3-x-4}{x+4}>0\)
\(\Leftrightarrow-\frac{1}{x+4}>0\)
\(\Leftrightarrow x+4< 0\)
\(\Leftrightarrow x< -4\)
Vậy \(x< -4\)
c) \(\frac{x+3}{2010}+\frac{x+2}{2011}+\frac{x+1}{2012}+\frac{x+2025}{4}=0\)
\(\Leftrightarrow\left(\frac{x+3}{2010}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2025}{4}-3\right)=0\)
\(\Leftrightarrow\frac{x+2013}{2010}+\frac{x+2013}{2011}+\frac{x+2013}{2012}+\frac{x+2013}{4}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2013=0\) (Vì \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\ne0\))
\(\Leftrightarrow x=-2013\)
Vậy \(x=-2013\)
Nhớ tick đó ✔✔✔
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 - 5 + 8 - .... - 101 ( 34 số hạng )
A = ( 2 - 5 ) + ( 8 - 11 ) + ( 14 -17 ) + .... + ( 98 - 101 ) ( 17 nhóm )
A = - 3 - 3 - ... - 3 ( 17 số hạng )
A = -3.17 = -51
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
I x + 1 I + I x + 2 I = 5
I x + 1 + x + 2 I = 5
I 2x + 3 I = 5
\(2x+3\hept{\begin{cases}-5\\5\end{cases}}\)
Khi 2x + 3 = -5
Thì ta có x = -4
Khi 2x + 3 = 5
Thì ta có x = 1