(c+1)+(x2)+(x+3)+...+(x+30)= 525
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2009<5*x -3 <2013
=> 2012<5*x<2016
=> 402<x<404=>x=403
b) 5*x +1+2+3+....+30=525
=> 5*x +31*30/2=525
=>5*x=60
=>x=12
5x-3=2010
5x-3=2011
5x-3=2012
5x=2007
5x=2008
5x=2009
ko có x thỏa mãn là số tự nhiên
b, 5x+ 465=525
5x=525-465
5x=60
x=60:5
x=14
![](https://rs.olm.vn/images/avt/0.png?1311)
525 + 525 x 3 + 998 x 525 - 525 x 2
= 525 x 4 + 523950 - 1050
= 2100 + 522900
= 525000
525 + 525 x 3 + 998 x 525 - 525 x 2
= 525 x 1 + 525 x 3 + 998 x 525 - 525 x 2
= 525 x (1 + 3 + 998 - 2)
=525 x 1000
=525000
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài toán :
Lời giải:
Tập xác định của hàm số
Giao điểm với trục hoành (OX)
Giao điểm với trục tung (OY)
Giới hạn hàm số tại vô cực
Khảo sát tính chẵn lẻ của hàm số
Giá trị của đạo hàm
Đạo hàm bằng 0 tại
Hàm số tăng trên
Hàm số giảm trên
Giá trị nhỏ nhất của hàm số
Trả lời:
\(A=525\times4+52\div4-4\times\left(40-15\right)-12+5\times2\)
\(A=525\times4+13-4\times25-12+10\)
\(A=\left(525-25\right)\times4+11\)
\(A=500\times4+11\)
\(A=2000+11\)
\(A=2011\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(x^3-\dfrac{1}{9}=0\)
\(\Rightarrow x^3-\left(\dfrac{1}{3}\right)^3=0\)
\(\Rightarrow\left(x-\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=0\\x^2+\dfrac{1}{3}x+\dfrac{1}{9}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x^2+\dfrac{1}{3}x=-\dfrac{1}{9}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)
1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)
2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)
3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+3\right)\left(1-x\right)>0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0.\\1-x>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0.\\1-x< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3.\\x< 1.\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3.\\x>1.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow-3< x< 1.\)
\(\left(x^2-1\right)\left(x^2-4\right)< 0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1< 0.\\x^2-4>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-1>0.\\x^2-4< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2< 1.\\x^2>4.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2>1.\\x^2< 4.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1.\\x>-1.\end{matrix}\right.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\\left[{}\begin{matrix}x< 2.\\x>-2.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-1< x< 1.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\-2< x< 2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2.\\x< -2.\\-2< x< -1.\\1< x< 2.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -2.\\x>2.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>14x+20+5=6x-9-9x
=>14x+25=-3x-9
=>17x=-34
=>x=-2
b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)
=>-28x-30=-50
=>-28x=-20
=>x=20/28=5/7
c: =>2x+x^3-x=x^3+1
=>x=1
d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22
=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22
=>-5x^2+2x-1+2x^2+x+22=0
=>-3x^2+3x+21=0
=>x^2-x-7=0
=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 30 ) = 525
( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 30 ) = 525
x . 30 + 465 = 525
x . 30 = 60
x = 2