Giải phương trình:
a.2x+x-3=5
b.3(2-x)-(x+2)=0
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
I
1
15 tháng 10 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x+6>=0\\x-2>=0\end{matrix}\right.\Leftrightarrow x>=2\)
\(\sqrt{x+6}-\sqrt{x-2}=2\)
=>\(\left(\sqrt{x+6}-\sqrt{x-2}\right)^2=4\)
=>\(x+6+x-2-2\sqrt{\left(x+6\right)\left(x-2\right)}=4\)
=>\(2\sqrt{\left(x+6\right)\left(x-2\right)}=2x+4-4=2x\)
=>\(\sqrt{\left(x+6\right)\left(x-2\right)}=x\)
=>\(\left\{{}\begin{matrix}x>=0\\\left(x+6\right)\left(x-2\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4x-12=x^2\end{matrix}\right.\)
=>x=3
b: ĐKXĐ: \(x-3>=0\)
=>x>=3
\(2\sqrt{x-3}-2x+3=0\)
=>\(\sqrt{4x-12}=2x-3\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\4x-12=4x^2-12x+9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\4x^2-12x+9-4x+12=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\4x^2-16x+21=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
ML
0
HN
1
LD
6 tháng 4 2022
\(a,\left|2x+2\right|+10=2x\)
*TH1 : \(\left|2x+2\right|=2x+2\Leftrightarrow2x+2>0\Leftrightarrow x>-1\)
\(\Rightarrow2x+2+10=2x\)
\(\Leftrightarrow2x-2x=-10-2\)
\(\Leftrightarrow0x=-12\left(vô\cdot lý\right)\)
*TH2 :\(\left|2x+2\right|=-2x-2\Leftrightarrow-2x-2< 0\Leftrightarrow x>-1\)
\(\Rightarrow-2x-2+10=2x\)
\(\Leftrightarrow-2x-2x=-10+2\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(nhận\right)\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
\(b,\left|x-6\right|=\left|3-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=3-2x\\x-6=-3+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-3;3\right\}\)
HN
3
DT
16 tháng 2 2022
a/
\(\Leftrightarrow x^2-2x+4-4=0\\ \Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow x=0;x-2=0\)
\(\Leftrightarrow x=0;x=2\)
DT
16 tháng 2 2022
b/
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3-2x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3-x\right)=0\)
\(\Rightarrow x=3\)
LV
1
13 tháng 3 2022
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
NH
2
26 tháng 1 2022
\(a,\left(đk:x\ge0\right)\)
\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)
\(x>0\)
\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)
\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)
\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)
\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)
S
3
22 tháng 2 2022
a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)
=>x=0
b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow x^2-4x+3=2x\left(x-3\right)-2\left(x^2-1\right)\)
\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)
\(\Leftrightarrow x^2+2x+1=0\)
=>x=-1(nhận)
22 tháng 2 2022
\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)
\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow-x^2-2x-1=0\)
\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)
17 tháng 6 2023
1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
a: =>3x=8
hay x=8/3
b: =>6-3x-x-2=0
=>-4x+4=0
hay x=1
\(a,2x+x-3=5\\ \Rightarrow3x=5+3\\ \Rightarrow3x=8\\ \Rightarrow x=\dfrac{8}{3}\\ b,3\left(2-x\right)-\left(x+2\right)=0\\ \Rightarrow6-3x-x-2=0\\ \Rightarrow4-4x=0\\ \Rightarrow4x=4\\ \Rightarrow x=1\)