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=> 3x . 31 + 3x . 32 = 92 . 4
=> 3x . ( 31 + 32 ) = 81.4
=> 3x . 12 = 324
=> 3x = 324 : 12
=> 3x = 27
=> 3x = 33
=> x = 3
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1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
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\(=\dfrac{6x^4-2x^3+5x^2-2}{3x^2-x+1}\)
\(=\dfrac{6x^4-2x^3+2x^2+3x^2-x+1+x-3}{3x^2-x+1}\)
\(=2x^2+1+\dfrac{x-3}{3x^2-x+1}\)
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\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=13/54
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`A(x)=0`
`<=>4x(x-1)-3x+3=0`
`<=>4x(x-1)-3(x-1)=0`
`<=>(x-1)(4x-3)=0`
`<=>` $\left[ \begin{array}{l}x=1\\x=\dfrac341\end{array} \right.$
`B(x)=0`
`<=>2/3x^2+x=0`
`<=>x(2/3x+1)=0`
`<=>` $\left[ \begin{array}{l}x=0\\x=-\dfrac32\end{array} \right.$
`C(x)=0`
`<=>2x^2-9x+4=0`
`<=>2x^2-8x-x+4=0`
`<=>2x(x-4)-(x-4)=0`
`<=>(x-4)(2x-1)=0`
`<=>` $\left[ \begin{array}{l}x=4\\x=\dfrac12\end{array} \right.$
![](https://rs.olm.vn/images/avt/0.png?1311)
4(x+3)-2(7-3x)=-3
<=> 4x + 12 - 14 + 6x = -3
<=> 4x + 6x = -3 - 12 + 14
<=> 10x = -1
<=> x = \(-\frac{1}{10}\)
\(4\left(x+3\right)-2\left(7-3x\right)=-3\)
\(4x+12-14+6x=-3\)
\(10x-2=-3\)
\(10x=-3+2\)
\(10x=1\)
\(x=1:10\)
\(x=\frac{1}{10}\)
\(\Leftrightarrow3^x\cdot82=3^{50}+3^{54}=3^{50}\cdot82\)
hay x=50