Ta thấy :   \(x^2+1\ge1\)  nên để   \(\left(3x-1\right)\left(x^2+1\right)< 0\)\(thì\) \(3x-1< 0\)\(hay\)  \(x< \frac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=x+1\\x\left(x+1\right)=-\left(x+1\right)\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left(x+1\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(\dfrac{7x+11}{2}-\dfrac{5x-3}{4}=\dfrac{x-6}{8}+\dfrac{3+x}{16} \)

⇔\(8\left(7x+11\right)-4\left(5x-3\right)=2\left(x-6\right)+\left(x+3\right)\)

⇔\(56x+88-20x+12=2x-12+x+3\)

⇔\(56x-20x-2x-x=-12+3-88-12\)

⇔\(33x=-109\)

⇔\(x=\dfrac{-109}{33}\)

               x + 1 = ( x + 1 )² 

               x + 1 = x² + 2x + 1

               x - 2x - x² = - 1 + 1

               - x - x² = 0

                    - x ( x + 1) = 0

          TH1: - x = 0 suy ra x = 0

          TH2: x + 1 = 0 suy ra x = - 1

               Vậy x = 0 hoặc x = - 1.

x = 0 nha!

 chúc bn học tốt~

\(x^2-5x+6=\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

1, <=>x^2-x-2 = x^2-4

<=>x^2-4-x^2+x+2 = 0

<=> x-2 = 0

<=> x=2

2, <=> (x-2).(x-3)=0

<=> x-2 = 0 hoặc x-3 = 0

<=> x=2 hoặc x=3

= x^3 + 2^3

= x^3 + 8

\(=x^3+8\)

a ) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x^2-3x^2\right)+\left(6x+3x\right)+\left(8-1\right)=17\)

\(\Leftrightarrow9x+7=17\)

\(\Leftrightarrow9x=10\)

\(\Leftrightarrow x=\dfrac{10}{9}\)

Vậy nghiệm của p/t là : \(\dfrac{10}{9}\)

b ) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=-\dfrac{11}{25}\)

Vậy nghiệm của p/t là : \(-\dfrac{11}{25}\)

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