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a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
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\(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-2\end{cases}}\)
\(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{-4}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : \(\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}4x=-2\\x^2=-1\left(loai\right)\end{cases}\Leftrightarrow}x=-2}\)
\(\left(3x+2\right).\left(x^2-1\right)=\left[\left(3x\right)^2-2^2\right].\left(x+1\right)\)
\(\Rightarrow\left(3x+2\right).\left(x-1\right).\left(x+1\right)-\left(3x-2\right).\left(3x+2\right).\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right).\left(x+1\right).\left[x-1-3x+2\right]=0\)
\(\Rightarrow\left(3x+2\right).\left(x+1\right).\left(-2x+1\right)=0\)
đến đây dễ rồi :))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9x^2-1+\left(3x-1\right).\left(x+2\right)=0\)
\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)
\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)
\(\Leftrightarrow12x^2+5x=3\)
\(\Leftrightarrow12x^2+5x-3=0\)
\(\Leftrightarrow12x^2-4x+9x-3=0\)
\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)
\(\Leftrightarrow\left(4x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,(x+3)(x^2+2021)=0`
`x^2+2021>=2021>0`
`=>x+3=0`
`=>x=-3`
`2,x(x-3)+3(x-3)=0`
`=>(x-3)(x+3)=0`
`=>x=+-3`
`b,x^2-9+(x+3)(3-2x)=0`
`=>(x-3)(x+3)+(x+3)(3-2x)=0`
`=>(x+3)(-x)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
`d,3x^2+3x=0`
`=>3x(x+1)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$
`e,x^2-4x+4=4`
`=>x^2-4x=0`
`=>x(x-4)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)
=> S={-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
=>(3x+1)(3x-1-4x-1)=0
=>(3x+1)(x+2)=0
=>x=-1/3 hoặc x=-2
\(9x^2-1=\left(3x+1\right)\left(4x+1\right)< =>\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(4x+1\right)< =>\left(3x+1\right)\left(3x-1-4x-1\right)=0< =>\left(3x+1\right)\left(-x-2\right)=0< =>\left[{}\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy .......