$\displaystyle \begin{array}{{>{\displaystyle}l}} 1.\ \\ a.\ x-\sqrt{4x-3} =2\\ ĐK\ x\geqslant \frac{3}{4}\\ \Leftrightarrow x-2=\sqrt{4x-3}\\ \Leftrightarrow x^{2} -4x+4=4x-3\ ( x\geqslant 2\\ \Leftrightarrow x^{2} -8x+7=0\\ \Leftrightarrow x=7\ ( chọn) ;\ x=1\ ( loại)\\ Vậy\ \ S=\{7\}\\ b.\ \sqrt{x+9} =5-\sqrt{2x+4}\\ ĐK\ x\geqslant -2\\ \Leftrightarrow \sqrt{x+9} +\sqrt{2x+4} =5\\ \Leftrightarrow x+9+2x+4+2\sqrt{2x^{2} +22x+36} =25\\ \Leftrightarrow \sqrt{2x^{2} +22x+36} =-\frac{3}{2} x+6\\ \Leftrightarrow 2x^{2} +22x+36=\frac{9}{4} x^{2} -18x+36\ ( x\leqslant 4)\\ \Leftrightarrow -\frac{1}{4} x^{2} +40x=0\\ \Leftrightarrow x=0\ ( TM) ;\ x=160\ ( loại)\\ Vậy\ S=\{0\}\\ c.\ \sqrt{3x+4} -\sqrt{2x+1} =\sqrt{x+3}\\ ĐK:\ x\geqslant -\frac{1}{2}\\ \Leftrightarrow \sqrt{3x+4} =\sqrt{x+3} +\sqrt{2x+1}\\ \Leftrightarrow 3x+4=x+3+2x+1+2\sqrt{2x^{2} +7x+3}\\ \Leftrightarrow \sqrt{2x^{2} +7x+3} =0\\ \Leftrightarrow 2x^{2} +7x+3=0\\ \Leftrightarrow x=-\frac{1}{2} \ ( TM) ;\ x=-3\ ( loại)\\ Vậy\ S=\left\{-\frac{1}{2}\right\}\\ d.\ \sqrt{x} -\sqrt{x-1} -\sqrt{x-4} +\sqrt{x+9} =0\\ ĐK:\ x\geqslant 1\\ \Leftrightarrow \ \sqrt{x} +\sqrt{x+9} =\sqrt{x-1} +\sqrt{x-4}\\ \Leftrightarrow x+x+9+2\sqrt{x^{2} +9x} =x-1+x-4+2\sqrt{x^{2} -5x+4}\\ \Leftrightarrow 9+2\sqrt{x^{2} +9x} =-5+2\sqrt{x^{2} -5x+4}\\ \Leftrightarrow 7+\sqrt{x^{2} +9x} =\sqrt{x^{2} -5x+4}\\ \Leftrightarrow 49+x^{2} +9x+14\sqrt{x^{2} +9x} =x^{2} -5x+4\\ \Leftrightarrow 14\sqrt{x^{2} +9x} =-14x-45\ \left( x\leqslant -\frac{45}{14}\right)\\ \Leftrightarrow 196\left( x^{2} +9x\right) =196x^{2} +2025+1260x\\ \Leftrightarrow 504x=2025\\ \Leftrightarrow x=\frac{225}{56} \ ( loại)\\ Vậy\ S=\{\emptyset \}\\ 2.\\ a.\ ĐK:\ x\geqslant 0\\ P=\left(\frac{6x+4}{\left(\sqrt{3x}\right)^{2} -2^{3}} -\frac{\sqrt{3x}}{3x+2\sqrt{3x} +4}\right) :\left(\frac{1+\left(\sqrt{3x}\right)^{3} -\sqrt{3x}\left( 1+\sqrt{3x}\right)}{1+\sqrt{3x}}\right)\\ P=\frac{6x+4-\left(\sqrt{3x} -2\right)\sqrt{3x}}{\left(\sqrt{3x} -2\right)\left( 3x+2\sqrt{3x} +4\right)} .\frac{1+\sqrt{3x}}{\left(\sqrt{3x} -1\right)^{2}\left(\sqrt{3x} +1\right)}\\ P=\frac{3x+4+2\sqrt{3x}}{\left(\sqrt{3x} -2\right)\left( 3x+2\sqrt{3x} +4\right)} .\frac{1+\sqrt{3x}}{\left(\sqrt{3x} -1\right)^{2}\left(\sqrt{3x} +1\right)}\\ P=\frac{3x+4+2\sqrt{3x}}{\left(\sqrt{3x} -2\right)\left( 3x+2\sqrt{3x} +4\right)} .\frac{1+\sqrt{3x}}{\left(\sqrt{3x} -1\right)^{2}\left(\sqrt{3x} +1\right)}\\ P=\frac{1}{\left(\sqrt{3x} -2\right)\left(\sqrt{3x} -1\right)} =\frac{1}{3x-3\sqrt{3x} +2}\\ b.\ P\in \mathbb{Z} \Leftrightarrow 3x-3\sqrt{3x} +2=\pm 1\\ \Leftrightarrow \sqrt{3x} =\frac{3\pm \sqrt{5}}{2}\\ \Leftrightarrow x=\frac{7\pm 3\sqrt{5}}{6} \ ( TM)\\ Vấy\ x=\left\{\frac{7\pm 3\sqrt{5}}{6}\right\}\\ 3.\ sin\alpha +cos\alpha =\frac{7}{5} \Rightarrow sin\alpha =\frac{7}{5} -cos\alpha \\ Ta\ có\ :\ \ sin^{2} \alpha +cos^{2} \alpha =1\Leftrightarrow \left(\frac{7}{5} -cos\alpha \right)^{2} +cos^{2} \alpha =1\\ \Leftrightarrow cos\alpha =\pm \frac{4}{5} \ mà\ 0^{o} < \alpha < 90^{o} \Rightarrow cos\alpha =\frac{4}{5}\\ \Rightarrow sin\alpha =\frac{3}{5} \Rightarrow tan\alpha =\frac{3}{4}\\ 4.\ A=\frac{x^{2} -5x+5}{x-x^{2}}\\ Ta\ có\ Ax-Ax^{2} =x^{2} -5x+5\ ( *)\\ \Leftrightarrow x^{2}( A+1) -x( A+5) +5=0\\ Để\ ( *) \ có\ nghiệm\ \ \Leftrightarrow ( A+5)^{2} -20( A+1) =A^{2} -10A+5\geqslant 0\\ \Leftrightarrow A\geqslant 5+2\sqrt{5}\\ Vậy\ GTNN\ A=5+2\sqrt{5}\\ 5.\ \sqrt{x} +2\sqrt{x+3} =x+4\\ ĐK:\ x\geqslant 0\\ \Leftrightarrow x+4x+12+4\sqrt{x^{2} +3x} =x^{2} +8x+16\\ \Leftrightarrow x^{2} +3x-4\sqrt{x^{2} +3x} +4=0\\ \Leftrightarrow \sqrt{x^{2} +3x} =2\\ \Leftrightarrow x^{2} +3x=4\\ \Leftrightarrow x=1\ ( TM) ;\ x=-4\ ( loại)\\ Vậy\ S=\{1\} \end{array}$
