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`(3x+2)*(x+4)-(3x-1)*(x-5)=0`

\(\Leftrightarrow3x\left(x+4\right)+2\left(x+4\right)-3x\left(x-5\right)+1\left(x-5\right)=0\)

\(\Leftrightarrow3x^2+3x\cdot4+2x+2\cdot4-3x^2+3x\cdot5+x-5=0\)

\(\Leftrightarrow3x^2+12x+2x+8-3x^2+15x+x-5=0\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(12x+2x+15x+x\right)+\left(8-5\right)=0\)

\(\Leftrightarrow30x+3=0\)

\(\Leftrightarrow30x=0-3\)

`=> 30x=-3`

`-> x=-3 \div 30`

`-> x=-1/10 `

\(\dfrac{x+1}{x-1}+\dfrac{x-2}{x+2}+\dfrac{x-3}{x+3}+\dfrac{x+4}{x-4}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x-4\right)+\left(x-2\right)\left(x-1\right)\left(x+3\right)\left(x-4\right)+\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x-4\right)+\left(x+4\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow4x^4+20x-96=0\)

\(\Leftrightarrow4\left(x^4+5x-24\right)=0\)

\(\Leftrightarrow x^4+5x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2,45...\\x=1,94...\end{matrix}\right.\)

Vậy: \(S=\left\{-2,45...;1,94...\right\}\)

  \(\left(x-5\right)\left(x-7\right)=0\)

=>\(\left\{{}\begin{matrix}x-5=0\\x-7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)

Vậy S ={5,7}

x - 5 = 0

x = 0 + 5

x = 5

⇔ x ∈ {5;7}

x - 5 = 0

x = 0 + 5

x = 5

⇔ x ∈ {5;7}

A+B>0

=>2*căn x-1>0

=>x>1/4

Gửi bạn

<=>  x^3 - 5x^2 +3x -4 =0

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