tìm x,biết a)x+1-1/2=-1/3giúp tôi với
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)
Bài 2:
a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=6\)
hay x=3
b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)
\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)
\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))
\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)
\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)
\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)
\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)
\(A=\dfrac{-5}{x-3}\)
b) Ta có: \(\left|x\right|=1\)
TH1: \(\left|x\right|=-x\) với \(x< 0\)
Pt trở thành:
\(-x=1\) (ĐK: \(x< 0\))
\(\Leftrightarrow x=-1\left(tm\right)\)
Thay \(x=-1\) vào A ta có:
\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)
TH2: \(\left|x\right|=x\) với \(x\ge0\)
Pt trở thành:
\(x=1\left(tm\right)\) (ĐK: \(x\ge0\))
Thay \(x=1\) vào A ta có:
\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)
c) \(A=\dfrac{1}{2}\) khi:
\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10=x-3\)
\(\Leftrightarrow x=-10+3\)
\(\Leftrightarrow x=-7\left(tm\right)\)
d) \(A\) nguyên khi:
\(\dfrac{-5}{x-3}\) nguyên
\(\Rightarrow x-3\inƯ\left(-5\right)\)
\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)
a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)
\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)
\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)
\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)
b: |x|=1
=>x=-1(loại) hoặc x=1(nhận)
Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)
c: A=1/2
=>x-3=-10
=>x=-7
d: A nguyên
=>-5 chia hết cho x-3
=>x-3 thuộc {1;-1;5;-5}
=>x thuộc {4;2;8;-2}
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, 1/2 +x=3/4`
`=> x= 3/4 -1/2`
`=> x= 3/4-2/4`
`=>x= 1/4`
`b, 5/2 -x=1/3`
`=> x= 5/2 -1/3`
`=> x= 15/6 - 2/6`
`=>x= 13/6`
`c, 2 . (1/3 +x)=1/5`
`=> 1/3 +x=1/5:2`
`=> 1/3 +x= 1/10`
`=>x= 1/10-1/3`
`=>x= 3/30 - 10/30`
`=>x=-7/30`
`d, 2/3 - (1/2 -x)=1/5`
`=> 1/2-x= 2/3 -1/5`
`=>1/2-x= 10/15 - 3/15`
`=>1/2-x=7/15`
`=>x= 1/2-7/15`
`=>x=1/30`
`1/2 + x = 3/4`
`=> x = 3/4 - 1/2`
`=> x = 1/4`
`5/2 - x = 1/3`
`=> x = 5/2 - 1/3`
`=> x = 13/6`
`2.(1/3 + x) = 1/5`
`=>1/3 + x = 1/10 `
`=> x = 1/10 - 1/3`
`=> x = -7/30`
`2/3 - (1/2 -x)= 1/5`
`=> 1/2 - x = 7/15`
`=> x = 1/2 - 7/15`
`=> x = 1/30`
![](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)
ĐKXĐ: x khác 0
1/(1 ×2)+ 1/(2 ×3)+ 1/(3 ×4 )+⋯.+ 1/(x ×( x+1))= 30/31
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/x +1 = 30/31
1 - 1/x+1 = 30/31
x+1-1/x+1 = 30/31
x/x+1 = 30/31
31x = 30x + 30
31x - 30x = 30
x = 30(TM)
Vậy x là 30.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\left(1+\dfrac{2}{3}\right)\cdot\left(1+\dfrac{2}{5}\right)\cdot\left(1+\dfrac{2}{7}\right)\cdot...\cdot\left(1+\dfrac{2}{2021}\right)\)
\(=\dfrac{5}{3}\cdot\dfrac{7}{5}\cdot\dfrac{9}{7}\cdot...\cdot\dfrac{2023}{2021}\)
\(=\dfrac{2023}{3}\)
\(a,x+1-\dfrac{1}{2}=\dfrac{-1}{3}\)
\(x+1=\dfrac{-1}{3}+\dfrac{1}{2}\)
\(x+1=\dfrac{-2}{6}+\dfrac{3}{6}\)
\(x+1=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}-1\)
\(x=\dfrac{1}{6}-\dfrac{6}{6}\)
\(x=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{-5}{6}\)
x + 1/2 = -1/3 x = -1/3 - 1/2 x = -5/6