Tìm x,biết:
a) \(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\) ; b) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\)
Trần Thiên KimNguyễn Huy TúPhương An
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HH
1
5 tháng 1 2022
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x^2+3x+1-2x^2-x+3=0\)
=>2x=-4
hay x=-2
16 tháng 7 2021
\(\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{1}{3}\\2x-\dfrac{2}{3}=\dfrac{-1}{3}\end{matrix}\right.\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
16 tháng 7 2021
a)\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
29 tháng 1 2022
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
21 tháng 2 2021
a)\(x\in R\)
b)\(x\ne1\)
c) \(x\notin\left\{1;2\right\}\)
d) \(x\notin\left\{3;-3\right\}\)
e) \(x\ne1\)
f) \(x\notin\left\{2;3\right\}\)
23 tháng 12 2021
\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)
8 tháng 7 2021
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
8 tháng 7 2021
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
19 tháng 6 2021
a) đk: x khác 1; \(\dfrac{3}{2}\)
\(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)
= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)
= \(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\dfrac{1-x}{-3x+5}=\dfrac{-1}{2x-3}\)
b) Có \(\left|3x-2\right|+1=5\)
<=> \(\left|3x-2\right|=4\)
<=> \(\left[{}\begin{matrix}3x-2=4< =>x=2\left(Tm\right)\\3x-2=-4< =>x=\dfrac{-2}{3}\left(Tm\right)\end{matrix}\right.\)
TH1: Thay x = 2 vào P, ta có:
P = \(\dfrac{-1}{2.2-3}=-1\)
TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:
P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)
c) Để P > 0
<=> \(\dfrac{-1}{2x-3}>0\)
<=> 2x - 3 <0
<=> x < \(\dfrac{3}{2}\) ( x khác 1)
d) P = \(\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)
<=> 2x - 3 = x2 - 6
<=> x2 - 2x - 3 = 0
<=> (x-3)(x+1) = 0
<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)
27 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)
\(=\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3\left(x-1\right)-2}{x-1}\)
\(=\dfrac{2x-5x+5}{2x-3}\cdot\dfrac{1}{3x-3-2}\)
\(=\dfrac{-3x+5}{2x-3}\cdot\dfrac{1}{3x-5}\)
\(=\dfrac{-1}{2x-3}\)
c) Để A>0 thì 2x-3<0
hay \(x< \dfrac{3}{2}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x\ne1\end{matrix}\right.\)
27 tháng 7 2021
Tham khảo:Cho biểu thức P= \((\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}):(3+\frac{2}{1-x})\) a) Rút gọn P b) Tính P với |3x-2|+1=5 c)... - Hoc24
a) ĐKXĐ:
Ta có:
a) \(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\Leftrightarrow\dfrac{2x+1}{x^2-2x+1}=\dfrac{2x+3}{x^2-1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-1\right)=\left(2x+3\right)\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=2x^2-2x+3x-3\)
\(\Leftrightarrow2x^2+2x+x+1-2x^2+2x-3x+3=0\)
\(\Leftrightarrow2x+4=0\Leftrightarrow2x=-4\Leftrightarrow x=\dfrac{-4}{2}=-2\) vậy \(x=-2\)
\(a.\)
\(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{2x+1}{x^2-2x+1}=\dfrac{2x+3}{x^2-1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-1\right)=\left(2x+3\right)\left(x^2-2x+1\right)\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)\left(x-1\right)=\left(2x+3\right)\left(x-1\right)^2\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)\)
\(\Rightarrow2x^2+2x+x+1=2x^2-2x+3x-3\)
\(\Rightarrow2x^2+2x+x+1-2x^2+2x-3x+3=0\)
\(\Rightarrow2x+4=0\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)