1/(x+2)+5/(2-x)=2x-3/x2-4
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DT
12 tháng 4 2022
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đkxđ : x khác 2 , x khác -2.
<=> \(\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{x^2-4}=0\)
<=> \(\dfrac{1.\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=>\(x-2-5x-10-2x+3=0\)
<=> \(-6x-9=0\)
<=> \(x=-\dfrac{9}{6}=-\dfrac{3}{2}\left(nhận\right)\)
Vậy pt có nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\)
27 tháng 2 2023
a: =2/5-3/5+3/7=3/7-1/5
=15/35-7/35
=8/35
b: =>5/7:x=4/3
=>x=5/7:4/3=5/7*3/4=15/28
c: =>x-1/3=15/8:4/5=15/8*5/4=75/32
=>x=75/32+1/3=257/96
d: =>2x+1/8=2/7
=>2x=9/56
=>x=9/112
e: =>2x=10/3-5/4-3/4=10/3-2=4/3
=>x=2/3
C
27 tháng 2 2023
\(a,\dfrac{2}{5}+\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{3}{5}\\=\left(\dfrac{2}{5}-\dfrac{3}{5}\right)+\dfrac{3}{7}\\ =-\dfrac{1}{5}+\dfrac{3}{7}\\ =-\dfrac{7}{35}+\dfrac{15}{35}\\ =\dfrac{8}{35}\\ b,1-\dfrac{5}{7}:x=-\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=1-\left(-\dfrac{1}{3}\right)\\ =>\dfrac{5}{7}:x=1+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{3}{3}+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{4}{3}\\ =>x=\dfrac{5}{7}:\dfrac{4}{3}\\ =>x=\dfrac{5}{7}.\dfrac{3}{4}\\ =>x=\dfrac{15}{28}\\ c,\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)=\dfrac{15}{8}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}:\dfrac{4}{5}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}.\dfrac{5}{4}\\ =>x-\dfrac{1}{3}=\dfrac{75}{32}\\ =>x=\dfrac{75}{32}+\dfrac{1}{3}\\ =>x=\dfrac{257}{96}\)
\(d,\dfrac{2}{3}:\left(2x+\dfrac{1}{8}\right)=\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}:\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}.\dfrac{3}{7}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{7}\\ =>2x=\dfrac{2}{7}-\dfrac{1}{8}\\ =>2x=\dfrac{16}{56}-\dfrac{7}{56}\\ =>2x=\dfrac{9}{56}\\ =>x=\dfrac{9}{56}:2\\ =>x=\dfrac{9}{112}\\ e,2x+\dfrac{3}{4}=\dfrac{10}{3}-\dfrac{5}{4}\\ =>e,2x+\dfrac{3}{4}=\dfrac{40}{12}-\dfrac{15}{12}\\ =>2x+\dfrac{3}{4}=\dfrac{25}{12}\\ =>2x=\dfrac{25}{12}-\dfrac{3}{4}\\ =>2x=\dfrac{25}{12}-\dfrac{9}{12}\\ =>2x=\dfrac{16}{12}\\ =>2x=\dfrac{4}{3}\\ =>x=\dfrac{4}{3}:2\\ =>x=\dfrac{4}{6}\\ =>x=\dfrac{2}{3}\)
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
10 tháng 7 2022
1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
=>x=0
2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5=7\)
=>6x-12=7
=>6x=19
hay x=19/6
10 tháng 7 2022
1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)-3=-6\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=-3\)
\(\Leftrightarrow14x=-3\)
hay x=-3/14
2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5\)
=>4x-12=7
=>4x=19
hay x=19/4
TN
5 tháng 1 2017
\(5\left(x+2\right)-x^2-2x=0\)
\(\Rightarrow5\left(x+2\right)-\left(x^2+2x\right)=0\)
\(\Rightarrow5\left(x+2\right)-x\left(x+2\right)=0\)
\(\Rightarrow\left(5-x\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)
16 tháng 3 2023
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
Về học lại hằng đẳng thức nha .-.
\(\Leftrightarrow\dfrac{\left(x+2\right)+5}{2-x}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow-\left(x+2\right)+5\left(x+2\right)=2x-3\\ \Leftrightarrow6x+12-2x+3=0\\ \Leftrightarrow4x+15=0\\ \Leftrightarrow x=\dfrac{-15}{4}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-2-5\left(x+2\right)-2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x-2-5x-10-2x-3=0\)
\(\Leftrightarrow-6x-15=0\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{15}{6}\left(n\right)\)
Vậy \(S=\left\{-\dfrac{15}{6}\right\}\)