\(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}\\ 2\sqrt{6}=\sqrt{4}.\sqrt{6}=\sqrt{24}\\ 4\sqrt{2}=\sqrt{16}.\sqrt{2}=\sqrt{32}\)

\(Vì\ :\ \)\(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)

\(=>2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)

\(=>\ Sắp\ xếp\ theo\ thứ\ tự\ tăng\ dần :\) \(2\sqrt{6};\sqrt{29};4\sqrt{2};3\sqrt{5}\)

\(Ta\ thấy :\)

\(\left(x-3\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\\ =>\left(x-3\right)^2+\left(y-2\right)^2\ge0\forall x;y\)

\(Dấu''=''\ xảy\ ra\ khi :\)

\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.=>\left(x;y\right)=\left(3;2\right)\)

\(Vậy\ (x;y)=(3;2)\)

\(B=x+y+\dfrac{1}{x}+\dfrac{1}{y}\ge x+y+\dfrac{4}{x+y}\)

\(Đặt:\ x+y=t(t\le4/5)\)

\(Khi\ đó:\)\(B\ge t+\dfrac{4}{t}\\ =>B\ge\left(\dfrac{25}{4}t+\dfrac{4}{t}\right)-\dfrac{21}{4}t\ge2\sqrt{\dfrac{25}{4}t.\dfrac{4}{t}}-\dfrac{21}{4}t\ge2\sqrt{25}-\dfrac{21}{4}.\dfrac{4}{5}=2.5-\dfrac{21}{5}=\dfrac{29}{5}\)

\(Dấu''=''\ xảy\ ra\ khi :\) \(\left\{{}\begin{matrix}x=y>0\\\dfrac{25}{4}t=\dfrac{4}{t}\\t=x+y=\dfrac{4}{5}\end{matrix}\right.< =>x=y=\dfrac{2}{5}\)

\(Vậy\ GTNN\ của\ B\ là:29/5<=>x=y=2/5\)

\(\left(b\right):\left(2x-3\right)^2-\left(x+1\right)^2=0\\ < =>\left(2x-3+x+1\right)\left(2x-3-x-1\right)=0\\ < =>\left(3x-2\right)\left(x-4\right)=0\\ =>\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.=>S=\left\{\dfrac{2}{3};4\right\}\)

\(\left(c\right):\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=4\\ < =>x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=4\\ < =>-9x^2+27x+9x^2+18x+9=4\\ < =>45x=-5\\ < =>x=-\dfrac{1}{9}=>S=\left\{-\dfrac{1}{9}\right\}\)

\(\left(d\right):x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\\ < =>x\left(x^2-25\right)-\left(x^3+8\right)=17\\ < =>x^3-25x-x^3-8=17\\ < =>-25x=25\\ < =>x=-1=>S=\left\{-1\right\}\)

\(8x^2+10x-3=\left(8x^2+12x\right)-\left(2x+3\right)\\ =4x\left(2x+3\right)-\left(2x+3\right)\\ =\left(2x+3\right)\left(4x-1\right)\)

\(98-5\left(6+x\right)=28\\ =>5\left(6+x\right)=98-28=70\\ =>6+x=\dfrac{70}{5}=14\\ =>x=14-6=8\)

\(ĐK:x\in Z;x\ne1\)

\(\dfrac{3x+2}{x-1}=\dfrac{3\left(x-1\right)+5}{x-1}=3+\dfrac{5}{x-1}\)

\(Để\ 3x+2/x-1\in Z\) 

\(=>\dfrac{5}{x-1}\in Z\\ =>5⋮\left(x-1\right)\\ =>x-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\\ =>x\in\left\{0;2;-4;6\right\}\left(TMDK\right)\)

\(\left(a\right):9x^2-12x+4=\left(3x\right)^2-2.3x.2+2^2=\left(3x-2\right)^2\\ \left(b\right):25+10x+x^2=x^2+2.x.5+5^2=\left(x+5\right)^2\\ \left(c\right):36x^2-25=\left(6x\right)^2-5^2=\left(6x-5\right)\left(6x+5\right)\\ \left(d\right):x^3-3x^2y+3xy^2-1=\left(x-1\right)^3\)

\(\left(a\right):A=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1>0\forall x\left(DPCM\right)\)\(\left(b\right):A\ge1\) \(=>GTNN\ của\ A\ là :1\)

\(Dấu ''=''\ xảy\ ra \ khi:(x-2)^2=0<=>x=2\)

\(\left(a\right):x^2-64=0< =>\left(x-8\right)\left(x+8\right)=0\\ =>\left[{}\begin{matrix}x-8=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(\left(b\right):4x^2-4x+1=0< =>\left(2x-1\right)^2=0\\ < =>2x-1=0\\ < =>x=\dfrac{1}{2}\)

\(\left(c\right):9-6x+x^2=0< =>\left(3-x\right)^2=0\\ < =>3-x=0\\ < =>x=3\)