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\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)
Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
a) 5(x−1)−(6−2x)=8x−3
=>5x−5−6+2x=8x−3
=> −x=8
=> x=−8
Vậy phương trình có nghiệm là x = -8
b)
\(\begin{array}{l}\frac{{2{\rm{x}} - 1}}{3} - \frac{{5 - 3{\rm{x}}}}{2} = \frac{{x + 7}}{4}\\\frac{{4\left( {2{\rm{x}} - 1} \right)}}{{12}} - \frac{{6\left( {5 - 3{\rm{x}}} \right)}}{{12}} = \frac{{3\left( {x + 7} \right)}}{{12}}\\8{\rm{x}} - 4 - 30 + 18{\rm{x}} = 3{\rm{x}} + 21\\8{\rm{x + 18x}} - 3{\rm{x}} = 21 + 4 + 30\\23{\rm{x}} = 55\\x = \frac{{55}}{{23}}\end{array}\)
Vậy phương trình có nghiệm là \(x = \frac{{55}}{{23}}\)
\(\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1}{\left(x-3\right)\left(2x-1\right)}=\frac{2x+5}{\left(x-3\right)\left(2x-1\right)}\)
\(\frac{\left(x-3\right)\left(x+4\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=\frac{\left(2x+5\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}\)
\(\Rightarrow x^2+x-12+x^2-x-2=2x^2+x-10\Leftrightarrow x=-4\)
\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{2x-5}{2x^2-7x+3}-\frac{x+1}{2x^2-7x+3}\)
\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{x+4}{2x^2-7x+3}\)
TH1:\(x+4\ne0\)
\(\Rightarrow2x^2-5x+2=2x^2-7x+3\)
\(\Rightarrow-5x+2=-7x+3\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
TH2:\(x+4=0\)
\(\Rightarrow x=-4\)
`7x -(2x+3) =5(x-2)`
`<=> 7x-2x-3=5x-10`
`<=> 7x-2x-5x=-10+3`
`<=> 0x=-7` ( vô lí )
Vậy phương trình vô nghiệm
\(x+\dfrac{2x-1}{5}=3+\dfrac{3-x}{4}\\ \Leftrightarrow\dfrac{20x}{20}+\dfrac{4\left(2x-1\right)}{20}=\dfrac{3\cdot20}{20}+\dfrac{5\left(3-x\right)}{20}\\ \Leftrightarrow20x+8x-4=60+15-5x\)
`<=> 20x+8x +5x = 60+15+4`
`<=> 33x= 79`
`<=> x= 79/33`
Vậy \(S=\left\{\dfrac{79}{33}\right\}\)
a) 7x−(2x+3)=5(x−2)
7x−2x−3=5x−10
0x=−7 (không thỏa mãn điều kiện a≠0)
b) x + \(\frac{{2{\rm{x}} - 1}}{5}\)=3 + \(\frac{{3 - x}}{4}\)
\(\frac{{20{\rm{x}} + 4\left( {2{\rm{x}} - 1} \right)}}{{20}} = \frac{{15 + 5\left( {3 - x} \right)}}{{20}}\)
20x+4(2x−1)=60+5(3−x)
20x+8x−4=60+15−5x
20x+8x+5x=60+15+4
33x=79
\(x = \frac{{79}}{{33}}\)
Vậy nghiệm của phương trình là \(x = \frac{{79}}{{33}}\)