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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}
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: \(S=\left\{\dfrac{3}{5}\right\}\)
11 tháng 7 2023
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
23 tháng 5 2022
Câu 1:
a: x+2=0
nên x=-2
b: (x-3)(2x+8)=0
=>x-3=0 hoặc 2x+8=0
=>x=3 hoặc x=-4
DT
23 tháng 5 2022
a .
x + 2 = 0
=> x = 0 - 2 = -2
b ) .
<=> x - 3 = 0 ; 2x + 8 = 0
= > x = 3 ; x = -8/2 = -4
c ) .
ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5
5 tháng 3 2021
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
NT
1
ND
6 tháng 2 2018
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3-6x+12=5\)
\(\Leftrightarrow-5x+15=5\)
\(\Leftrightarrow x=2\)
NT
2
6 tháng 2 2018
điều kiện xác định : \(x\ne-2\)
ta có : \(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
\(\Leftrightarrow\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}-\dfrac{5}{x^2-2x+4}=0\)
\(\Leftrightarrow\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5}{x^2-2x+4}=0\)
\(\Leftrightarrow\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-4x+8}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5x+10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)\(\Leftrightarrow\dfrac{2x^2-4x+8-2x^2-16-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{-9x-18}{\left(x+2\right)\left(x^2-2x+4\right)}=0\Leftrightarrow-9x-18=0\)
\(\Leftrightarrow-9x=18\Leftrightarrow x=-2\left(loại\right)\)
vậy phương trình vô nghiệm
G
6 tháng 2 2018
Giải:
\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\) (1)
ĐKXĐ: \(x\ne-2\)
\(\left(1\right)\Leftrightarrow\dfrac{2\left(x^2-2x+4\right)}{x^3+8}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5\left(x+2\right)}{x^3+8}\)
\(\Rightarrow2\left(x^2-2x+4\right)-2x^2+16=5\left(x+2\right)\)
\(\Rightarrow2x^2-4x+8-2x^2+16=5x+10\)
\(\Rightarrow-4x-5x=10-8-16\)
\(\Rightarrow-9x=-14\)
\(\Rightarrow x=-\dfrac{14}{-9}=\dfrac{14}{9}\) (thoả mãn ĐKXĐ)
Vậy ...
MV
12 tháng 8 2017
2.
\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
\(\Rightarrow x>0\)
Vậy \(x>0\)
HV
1
1 tháng 5 2017
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\2-x\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Pt \(\Leftrightarrow\) \(\dfrac{\left(x-2\right)}{x^2-4}+\dfrac{-5\left(x+2\right)}{x^2-4}=\dfrac{2x-3}{x^2-4}\)
\(\Leftrightarrow x-2-5x-10=2x-3\)
\(\Leftrightarrow x-5x-2x=10+2-3\)
\(\Leftrightarrow-6x=9\)
\(\Leftrightarrow x=\dfrac{-3}{2}\) ( thỏa mãn)
Vậy nghiệm của pt là \(x=\dfrac{-3}{2}\)
\(\dfrac{x+2}{x-2}-\dfrac{x-3}{x+2}=5\\ \Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}=5\\ \Leftrightarrow\dfrac{x^2+4x+4-x^2+5x-6}{\left(x^2-4\right)}=5\\ \Leftrightarrow9x-2=5\left(x^2-4\right)\\ \Leftrightarrow5x^2-20=9x-2\\ \Leftrightarrow5x^2-9x-18=0\\ \Leftrightarrow\left(5x+6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+6=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{6}{5};3\right\}\)
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