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23 tháng 7 2021
Bài 1:
a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)
\(\Leftrightarrow6-8x-10+2x-5=0\)
\(\Leftrightarrow-6x+11=0\)
\(\Leftrightarrow-6x=-11\)
hay \(x=\dfrac{11}{6}\)
b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)
\(\Leftrightarrow6-12x-11+3x-1=0\)
\(\Leftrightarrow-9x-6=0\)
\(\Leftrightarrow-9x=6\)
hay \(x=-\dfrac{2}{3}\)
16 tháng 1 2022
\(\Leftrightarrow2y^2+2y-3y-3+y^2-2y=3y^2+12y+12\)
=>-3y-3=12y+12
=>-15y=15
hay y=-1
16 tháng 2 2022
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
DT
16 tháng 2 2022
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
LV
2
23 tháng 9 2023
\(2x^2-3x-5=0 \\ \Leftrightarrow2x^2+2x-5x-5=0\\ \Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=5\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\end{matrix}\right.\\ Vậy.S=\left\{\dfrac{5}{2};-1\right\}\)
T
23 tháng 9 2023
\(2x^2-3x-5=0\)
\(\Leftrightarrow2x^2+2x-5x-5=0\)
\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\end{matrix}\right.\)
Vậy \(x=\dfrac{5}{2};x=-1\) là các nghiệm của phương trình.
#\(Toru\)
VD
16 tháng 3 2022
\(a,3\left(x-1\right)-7=5\left(x+2\right)\\ \Leftrightarrow3x-3-7=5x+10\\ \Leftrightarrow3x-10=5x+10\\ \Leftrightarrow2x+20=0\\ \Leftrightarrow x=-10\\ b,\dfrac{3x+1}{2}-\dfrac{x+3}{5}=\dfrac{x}{10}+2\\ \Leftrightarrow\dfrac{5\left(3x+1\right)}{10}-\dfrac{2\left(x+3\right)}{10}-\dfrac{x}{10}-\dfrac{20}{10}\\ \Leftrightarrow15x+5-2x-6-x-20=0\\ \Leftrightarrow12x-21=0\\ \Leftrightarrow x=\dfrac{7}{4}\)
\(c,ĐKXĐ:x\ne\pm1\\ \dfrac{x-2}{x+1}-\dfrac{x}{x-1}=\dfrac{x-8}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-8}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2-3x+2-x^2-x-x+8}{\left(x+1\right)\left(x-1\right)}=0\\ \Rightarrow-5x+10=0\\ \Leftrightarrow x=2\left(tm\right)\)
16 tháng 3 2022
a,3x-3-7=5x+10
<=>3x-10-5x-10=0
<=>-2x-20=0
<=>x=-10
b,quy đồng mẫu là 10 , ta được
5(3x+1)-2(x+3)=x+20
<=>15x+5-2x-6=x+20
<=>13x-x-1-20=0
<=>12x=21
<=>x=\(\dfrac{7}{4}\)
c,đk:x khác 1,-1
quy đồng khử mẫu ta được
(x-2)(x-1)-x(x+1)=x-8
<=>x\(^2\) -3x+2-x\(^2\) -x=x-8
<=>-4x-x=-8-2
<=>-5x=-10
<=>x=2
PT
24 tháng 1 2022
a) \(\dfrac{2x-1}{15}-\dfrac{3-x}{3}=1\)
\(\Leftrightarrow2x-1-5\left(3-x\right)=15\)
\(\Leftrightarrow2x-1-15+5x-15=0\)
\(\Leftrightarrow x=\dfrac{31}{7}\)
b) \(\dfrac{17-2}{18}=2x+\dfrac{10-x}{6}\)
\(\Leftrightarrow17-2=36x+3\left(10-x\right)\)
\(\Leftrightarrow36x+30-3x-15=0\)
\(\Leftrightarrow x=\dfrac{-5}{11}\)
15 tháng 1 2022
e: \(\Leftrightarrow x^3-4x^2+4x-2x^2+2x-x^3+6x^2-6x+2022=0\)
=>0x+2022=0(vô lý)
15 tháng 1 2022
d)⇔\(\dfrac{4\left(x+1\right)+9\left(2x+1\right)}{12}=\dfrac{2.\left[2x+3\left(x+1\right)\right]+7+12x}{12}\)
⇔\(4\left(x+1\right)+9\left(2x+1\right)=2.\left[2x+3\left(x+1\right)\right]+7+12x\)
⇔4x+4+18x+9=4x+6x+6+7+12x
⇔22x+13=22x+13
⇔0=0 (đúng)
Vậy tập nghiệm của phương trình: S=R
TM
24 tháng 1 2022
1. \(1+\dfrac{2x-5}{6}=\dfrac{3-x}{4}\)
\(\Leftrightarrow\dfrac{4x+2}{12}=\dfrac{9-3x}{12}\)
\(\Leftrightarrow4x+2=9-3x\)
\(\Leftrightarrow7x=7\)
\(\Leftrightarrow x=1\)
Vậy: Phương trình có tập nghiệm \(S=\left\{1\right\}\)
2. \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow\dfrac{6x-18}{10}=\dfrac{15-5x}{10}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=23\)
\(\Leftrightarrow x=\dfrac{23}{11}\)
Vậy: .....
26 tháng 1 2022
1) \(3x-2=6.\Leftrightarrow x=\dfrac{8}{3}.\)
2) \(x\left(x+8\right)-12=x^2+6x.\Leftrightarrow x^2+8x-12-x^2-6x=0.\\ \Leftrightarrow2x=12.\Leftrightarrow x=6.\)
3) \(\dfrac{3x+1}{6}=\dfrac{5x-2}{9}.\Leftrightarrow27x+9=30x-12.\\ \Leftrightarrow-3x=-21.\Leftrightarrow x=7.\)
4) \(\left(x-2\right)\left(2x+3\right)-x^2+4=0.\\ \Leftrightarrow2x^2+3x-4x-6-x^2+4=0.\Leftrightarrow x^2-x-2=0.\Leftrightarrow\left(x-2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=-1.\end{matrix}\right.\)
=>(2x+5)[1/6-1/3(x-10)]=0
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\\dfrac{1}{3}\left(x-10\right)=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{21}{2}\end{matrix}\right.\)