Tìm x biết:|x-1|-x+2=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)
\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)
\(x-\dfrac{2}{3}\times x-6=1\)
\(x\times\left(1-\dfrac{2}{3}\right)=7\)
\(x\times\dfrac{1}{3}=7\)
\(x=21\)
\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)
\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)
\(15x-11=9+3x\)
\(12x=20\)
\(x=\dfrac{5}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
............................. Đấng Ed bảo ko chắc cho lắm nên sai thì sr nhé -,-
\(a)\)\(\left|x-1\right|+\left|x-2\right|+...+\left|x-8\right|=22\)
+) Với \(x\ge8\) ta có :
\(x-1+x-2+...+x-8=22\)
\(\Leftrightarrow\)\(8x-36=22\)
\(\Leftrightarrow\)\(x=\frac{29}{4}\)( không thỏa mãn )
+) Với \(x< 1\) ta có :
\(1-x+2-x+...+8-x=22\)
\(\Leftrightarrow\)\(36-8x=22\)
\(\Leftrightarrow\)\(x=\frac{7}{4}\) ( không thỏa mãn )
Vậy không có x thỏa mãn đề bài
\(b)\)\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-100\right|=2500\)
+) Với \(x\ge100\) ta có :
\(x-1+x-2+x-3+...+x-100=2500\)
\(\Leftrightarrow\)\(100x-5050=2500\)
\(\Leftrightarrow\)\(x=\frac{151}{2}\) ( không thỏa mãn )
+) Với \(x< 1\) ta có :
\(1-x+2-x+3-x+...+100-x=2500\)
\(\Leftrightarrow\)\(5050-100x=2500\)
\(\Leftrightarrow\)\(x=\frac{51}{2}\) ( không thỏa mãn )
Vậy không có x thỏa mãn đề bài
Bài 2 :
+) Với \(x\ge-1\) ta có :
\(x+1+x+2+...+x+100=605x\)
\(\Leftrightarrow\)\(100x+5050=605x\)
\(\Leftrightarrow\)\(x=10\) ( thỏa mãn )
+) Với \(x< -100\) ta có :
\(-x-1-x-2-...-x-100=605x\)
\(\Leftrightarrow\)\(-100x-5050=605x\)
\(\Leftrightarrow\)\(x=\frac{-1010}{141}\) ( không thỏa mãn )
Vậy \(x=10\)
~ Đấng phắn ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
ĐKXĐ: \(x\ne1\)
\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)
\(\Rightarrow C=\left[2x^2+1-1\left(x^2+x+1\right)\right]\div\left[\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2\right)\right]\)
\(\Rightarrow C=(2x^2+1-x^2-x-1)\div\left[\left(x-1\right)\left(x^2+x+1-x^2+2\right)\right]\)
\(\Rightarrow C=\left(x^2-x\right)\div\left[\left(x-1\right)\left(x+3\right)\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x = 1/8 - y/4 = (1-2y)/8
<=> x = 5*8/(1-2y) ; thấy 1-2y là số lẻ nên UCLN(8,1-2y) = 1
do đó x/8 = 5/(1-2y) (*)
x, y nguyên khi 1-2y phải là ước của 5
* 1-2y = -1 => y = 1 => x = -40
* 1-2y = 1 => y = 0 => x = 40
* 1-2y = -5 => y = 3 => x = -8
* 1-2y = 5 => y = -2 => x = 8
vậy có 4 cặp (x,y) nguyên (-40,1) ; (40, 0) ; (-8, -5) ; (8, 5) .
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: x<>0; x<>1
\(P=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |2x+1|=3
=>x=1(loại); x=-2(nhận)
Khi x=-2 thì P=4/-3=-4/3
c: P=-1/2
=>x^2/x-1=-1/2
=>2x^2=-x+1
=>2x^2+x-1=0
=>2x^2+2x-x-1=0
=>(x+1)(2x-1)=0
=>x=1/2; x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1-x}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{\left[x\left(x^4+x^2+1\right)\right]}\)
\(\Leftrightarrow\frac{\left(1-x\right)x\left(x^2-x+1\right)\left(x^4+x^2+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)}\)\(-\)\(\frac{x\left(x-1\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)}\)\(=\)\(\frac{3\left(x^2-x+1\right)\left(x^2+x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)}\)
\(\Rightarrow\left(1-x\right)x\left(x^2-x+1\right)\left(x^4+x^2+1\right)-x\left(x-1\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)=\)\(3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x-x^2\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)-\left(x^2-x\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)=\)\(\left(3x^2-3x+3\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x^3-x^2+x-x^4+x^3-x^2\right)\left(x^4+x^2+1\right)-\left(x^4+x^3+x^2-x^3-x^2-x\right)\left(x^4+x^2+1\right)=\) \(3x^4+3x^3+3x^2-3x^3-3x^2-3x+3x^2+3x+3\)
\(\Leftrightarrow\left(2x^3-2x^2+x-x^4\right)\left(x^4+x^2+1\right)-\left(x^4-x\right)\left(x^4+x+1\right)=3x^4+3x^2+3\)
\(\Leftrightarrow\left(x^4+x^2+1\right)\left(2x^3-2x^2+x-x^4-x^4+x\right)=3x^4+3x^2+3\)
\(\Leftrightarrow\left(x^4+x^2+1\right)\left(2x^3-2x^2+2x-2x^4\right)=3x^4+3x^2+3\)
\(\Leftrightarrow2x^7-2x^6+2x^5-2x^8+2x^5-2x^4+2x^3-2x+2x^3-2x^2+2x-2x^4-3x^4-3x^2-3=0\)
\(\Leftrightarrow2x^7-2x^6+4x^5-2x^8-7x^4+x^2-3=0\)
Đến đây thì chịu òi :^ Sr nha
\(\frac{1-x}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
Ta có \(x^4+x^2+1=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
=> \(\left(1-x\right)\left(\frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right)=\frac{3}{x\left(x^4+x^2+1\right)}\)
<=>\(\left(1-x\right)\left(2x^2+2\right).x=3\)
Do \(2x^2+2>0\)
=> \(\left(1-x\right).x>0\)
=> \(0< x< 1\)=> \(2x^2+2< 4\)
Pt<=> \(\left(x-x^2\right)\left(2x^2+2\right)=3\)
Mà \(x-x^2\le\frac{1}{4};2x^2+2< 4\)
=> \(VT< 1\)
=> PT vô nghiệm