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BPT\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)\left(x-1\right)\left(x^2+x+1\right)\left(3-x\right)\left(x+1\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(3-x\right)\left(x+1\right)\ge0\) VÌ \(\left(\left(x^2+3x+9\right).\left(x^2+x+1\right)>0với\forall x\right)\)
\(\Leftrightarrow\left(x-3\right)^2.\left(1-x\right)\left(1+x\right)\ge0\)
\(\Leftrightarrow\left(1-x\right)\left(1+x\right)\ge0\left(vì\left(x-3\right)^2\ge0voi\forall x\right)\)
\(\Leftrightarrow-1\le x\le1\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
\(a,3x-2\ge x+4\) => \(2x\ge6\)=>\(x\ge3\)
`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}
a. (x-3)(x\(^2\)+6x+9)(x-1)(x\(^2\)+2x+1)(-x\(^2\)+2x+3)=0
\(\Leftrightarrow\)(x-3)(x\(^2\)+6x+9)(x-1)(x\(^2\)+2x+1)(x-3)(x+1)=0
\(\left(x^3-27\right)\left(x^3-1\right)\left(2x+3-x^2\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)\left(x-1\right)\left(x^2+x+1\right)\left[4-\left(x-1\right)^2\right]\ge0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+\frac{3}{2}\right)^2+\frac{27}{4}\right]\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\left(4-x+1\right)\left(4+x-1\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(5-x\right)\left(x+3\right)\left[...\right]\left[...\right]\ge0\)(1)
Do [...] và [...] > 0
nên \(\left(1\right)\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(5-x\right)\left(x+3\right)\ge0\)
\(\Leftrightarrow\left(x-5\right)\left(x-3\right)\left(x-1\right)\left(x+3\right)\le0\)
Có: \(x-5< x-3< x-1< x+3\)
Nên xảy ra các trường hợp sau :
TH1:\(\hept{\begin{cases}x-5\le0\\x-3\ge0\end{cases}}\)(Tự giải)
TH2:\(\hept{\begin{cases}x-1\le0\\x+3\ge0\end{cases}}\)(Tự giải)
Cuối cùng gộp khoảng (Nếu được)
Kết luận......