tim x biết ( x + 2 ) = ( x - 2 ) ( x + 1 )
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)
1) \(M=\frac{x^2+y^2+7}{x^2+y^2+5}=1+\frac{2}{x^2+y^2+5}\)
Ta có: \(x^2+y^2\ge0,\forall x;y\)
=> \(x^2+y^2+5\ge5\) với mọi x; y
=> \(\frac{2}{x^2+y^2+5}\le\frac{2}{5}\)
=> \(M\le1+\frac{2}{5}=\frac{7}{5}\)
Dấu "=" xảy ra <=> x = y = 0
Vậy max M = 7/5 đạt tại x = y = 0
2) \(f\left(x-1\right)=x^2-3x+5=x^2-x-2x+2+3\)
\(=x\left(x-1\right)-2\left(x-1\right)+3=x\left(x-1\right)-\left(x-1\right)-\left(x-1\right)+3\)
\(=\left(x-1\right)\left(x-1\right)-\left(x-1\right)+3\)
=> \(f\left(x\right)=x.x-x+3=x^2-x+3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x + (x + 1) + (x + 2) + ... + (x + 2003) = 2004
=> x + x + 1 + x + 2 + .... + x + 2003 = 2004
=> 2004x + 2007006 = 2004
=> 2004x = 2005002
=> x = 1000,5
Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=2004\)
\(\Leftrightarrow2004x+2007006=2004\)
\(\Leftrightarrow2004x=-2005002\)
hay \(x=-\dfrac{2001}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
\(\frac{x+1}{x.\left(x+1\right)}-\frac{x}{x.\left(x+1\right)}=\frac{1}{2}\)
\(\frac{x+1-x}{x.\left(x+1\right)}=\frac{1}{2}\)
\(\frac{1}{x.\left(x+1\right)}=\frac{1}{1.2}\)
\(x.\left(x+1\right)=1.2\)
Vì x và x+1 là 2 số tự nhiên liên tiếp mà x<x+1
=>x=1, x+1=2
Vậy x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
(+) Th1: l x - 1 l = x - 1 khi x - 1 >= 0 => x >= 1
Thay vào pt ta có:
x - 1 = x + 2
0x = 3 ( loại )
(+) Th2 : l x - 1 l = 1 - x khi x - 1 <=0 => x <= 1
Thay vào ta có :
1 - x = x + 2
-x - x = 2 - 1
-2x = 1
x = -1/2 ( Tm)
Vậy x = -1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(5x-1\right)-x\left(x+1\right)+x^2=14\)
\(\Leftrightarrow15x-3-x^2-x+x^2=14\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(15x-x\right)-3=14\)
\(\Leftrightarrow14x=17\)
\(\Leftrightarrow x=\frac{17}{14}\)
Vậy \(x=\frac{17}{14}\)
3(5x - 1) - x(x+1)+x 2 = 14
➡️15x - 3 - x 2 - x + x 2 = 14
➡️(15x - x ) + ( -x 2 + x 2 ) - 3= 14
➡️14x -3 = 14
➡️14x = 14+3
➡️14x = 17
➡️x = 17/14
Hok tốt~
![](https://rs.olm.vn/images/avt/0.png?1311)
|x+1|+|x+2|+|x+3|+|x+4|+|x+5|=6x
=> 6x > 0 => x > 0 => x+1; x+2 ; x+ 3; x+ 4 ; x+ 5 > 0
=> |x+1| + |x+2|+ |x+3| + |x +4| + |x+5| = (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 5x + (1+2+3+4+5) = 5x + 15
=> 5x + 15 = 6x => 15 = 6x - 5x => 15 = x
Vậy x = 15