Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.x\in \left\{3;3\pm 2\sqrt{5}\right\}\\ b.x\in \left\{\frac{3}{2} ;2\right\} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \left( x^{2} -6x\right)^{2} -2( x-3)^{2} =81\\ \Leftrightarrow x^{4} -12x^{3} +36x^{2} -2x^{2} +12x-18-81=0\\ \Leftrightarrow x^{4} -12x^{3} +34x^{2} +12x-99=0\\ \Leftrightarrow x^{4} -3x^{3} -9x^{3} +27x^{2} +7x^{2} -21x+33x-99=0\\ \Leftrightarrow x^{3}( x-3) -9x^{2}( x-3) +7x( x-3) +33( x-3) =0\\ \Leftrightarrow \left( x^{3} -9x^{2} +7x+33\right)( x-3) =0\\ \Leftrightarrow x=3\ hoặc\ x^{3} -9x^{2} +7x+33=0( 1)\\ ( 1) \Leftrightarrow x=3\pm 2\sqrt{5}\\ Vậy\ x\in \left\{3;3\pm 2\sqrt{5}\right\}\\ b.\ 2x^{2} -7x+6=0\\ \Leftrightarrow 2x^{2} -4x-3x+6=0\\ \Leftrightarrow 2x( x-2) -3( x-2) =0\\ \Leftrightarrow ( 2x-3)( x-2) =0\\ \Leftrightarrow 2x-3=0\ hoặc\ x-2=0\\ \Leftrightarrow x=\frac{3}{2} \ hoặc\ x=2\\ Vậy\ x\in \left\{\frac{3}{2} ;2\right\} \end{array}$
