a) \(2(x-1)\)2 + \((x + 3)\)2 = \(3(x-2)(x+1)\)
⇔\(2x^2-4x+2+x^2+6x+9=3x^2+3x-6x-6\)
⇔\(2x^2+x^2-3x^2-4x+6x-3x+6x=-2-9-6\)
⇔\(5x=-17\)
⇔\(x=\frac{-17}{5}\)
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b: \(\Leftrightarrow x^2+4x+4-2x+6-x^2-2x-1=0\)
=>9=0(vô lý)
c: \(\Leftrightarrow x^2-2x+1+x^2-4x+4=2x^2+16x+32-22x-27\)
=>\(2x^2-6x+5-2x^2+6x-5=0\)
=>0x=0(luôn đúng)
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.
, ta có: 
.![(3-x)(2+x) \leq \dfrac{25}{4}, \, \forall x\in [-2; 3]](https://s0.wp.com/latex.php?latex=%283-x%29%282%2Bx%29+%5Cleq+%5Cdfrac%7B25%7D%7B4%7D%2C+%5C%2C+%5Cforall+x%5Cin+%5B-2%3B+3%5D&bg=ffffff&fg=4e4e4e&s=0 "(3-x)(2+x) \leq \dfrac{25}{4}, \, \forall x\in [-2; 3]")
.![(2-x).(1+2x) \leq \dfrac{25}{8}, \, \forall x\in [-\dfrac{1}{2}; 2]](https://s0.wp.com/latex.php?latex=%282-x%29.%281%2B2x%29+%5Cleq+%5Cdfrac%7B25%7D%7B8%7D%2C+%5C%2C+%5Cforall+x%5Cin+%5B-%5Cdfrac%7B1%7D%7B2%7D%3B+2%5D&bg=ffffff&fg=4e4e4e&s=0 "(2-x).(1+2x) \leq \dfrac{25}{8}, \, \forall x\in [-\dfrac{1}{2}; 2]")
