giải pt:│x+5│+│x+2│+│x-1│=6
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
15 tháng 4 2020
\(ĐKXĐ:x\ne5,8\)
\(\frac{6}{x-5}+\frac{x+2}{x-8}=\frac{18}{\left(x-5\right)\left(8-x\right)}-1\)
\(\Rightarrow\frac{6}{x-5}+\frac{x+2}{x-8}=-\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)
\(\Rightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)
\(\Rightarrow x^2+3x-58=-x^2+13x-58\)
\(\Rightarrow2x^2-10x=0\)
\(\Rightarrow2x\left(x-5\right)=0\)
\(\Rightarrow x\in\left\{0,5\right\}\)
GK
1
14 tháng 9 2023
\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)
\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)
\(\Leftrightarrow4x-2< 3x+18\)
\(\Leftrightarrow4x-3x< 2+18\)
\(\Leftrightarrow x< 20\)
\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)
\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)
\(\Leftrightarrow5x-11>4x+4\)
\(\Leftrightarrow5x-4x>11+4\)
\(\Leftrightarrow x>15\)
26 tháng 8 2022
\(\dfrac{1}{x-3}=\dfrac{x^2-3x+5}{x^2-x-6}\)
Suy ra: \(x^2-3x+5=x+2\)
=>x2-4x+3=0
=>(x-3)*(x-1)=0
=>x=1(nhận) hoặc x=3(loại)
NU
3
13 tháng 4 2021
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)
Vậy phương trình vô nghiệm
25 tháng 2 2021
`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`
`<=>(-x-1)/(x^2-2x-3)-(x-3)/(x^2-2x-3)=(x^2+x)/(x^2-2x-3)-(x-1)^2/(x^2-2x-3)`
`<=>-x-1-x+3=x^2+x-x^2+2x-1`
`<=>-2x+2=3x-1`
`<=>5x=3`
`<=>x=3/5`
Vậy `S={3/5}`
`1/(x-2)-6/(x+3)=6/(6-x^2-x)(x ne 2,-3)`
`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`
`<=>x+3-6x+12+6=0`
`<=>-5x+21=0`
`<=>x=21/5`
Vậy `S={21/5}`
25 tháng 2 2021
a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
Ta có: \(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Leftrightarrow\dfrac{-1\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{x-3}{\left(x+1\right)\left(x-3\right)}=\dfrac{x\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-3\right)\left(x+1\right)}\)
Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)
\(\Leftrightarrow3x-1=-2x+2\)
\(\Leftrightarrow3x+2x=2+1\)
\(\Leftrightarrow5x=3\)
hay \(x=\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{\dfrac{3}{5}\right\}\)
NT
1
ND
6 tháng 2 2018
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3-6x+12=5\)
\(\Leftrightarrow-5x+15=5\)
\(\Leftrightarrow x=2\)
Ta có:\(\left|x+5\right|+\left|x-1\right|=\left|x+5\right|+\left|1-x\right|\ge\left|x+5+1-x\right|=6\)
\(\Rightarrow\left|x+5\right|+\left|x+2\right|+\left|x-1\right|=6\Leftrightarrow x=-2\)
* với x ≥ -5
-3x-6=6
⇔ -3x=12
⇔ x=-4 (tm)
*với -5 ≤ x < -2
-x+4=6
⇔ -x=2
⇔ x=-2 (ktm)
* với -2 ≤ x < 1
x+8=6
⇔ x=6-8
⇔ x= -2 (tm)
* với x< 1
3x+4 =6
⇔ 3x=2
⇔ x= \(\dfrac{2}{3}\) (tm)
vậy tập nghiệm của phương trình là S \(\left\{\dfrac{2}{3};-2;-4\right\}\)