Giải hệ:
[tex]\left\{\begin{matrix} \sqrt{5x^2+10xy+10y^2}+ \sqrt{5x^2 -8xy+4y^2}= 3(x+y)\\ (8x-6)\sqrt{y-1} = (2+\sqrt{y-2})(x+3+4\sqrt{x-2}) \end{matrix}\right.[/tex]
[tex]\sqrt{5x^2+10xy+10y^2}+ \sqrt{5x^2 -8xy+4y^2}\geq (2x+3y)+ x=3(x+y)\\\\ "=" <=> x=y[/tex]
khi đó: [tex](8x-6)\sqrt{y-1} = (2+\sqrt{y-2})(x+3+4\sqrt{x-2})\\\\ <=> (8x-6)\sqrt{x-1} = (2+\sqrt{x-2})(x+3+4\sqrt{x-2})\\\\ <=> (8x-6)\sqrt{x-1} = 6x-2+(x+11).\sqrt{x-2}\\\\ <=> (8x-6).[\sqrt{x-1}-(\frac{3x+2}{8})]=(x+11).[\sqrt{x-2}-(\frac{3x-6}{4})]+6x-2-(8x-6).(\frac{3x+2}{8})+(x+11).(\frac{3x-6}{4})\\\\ <=> (4x-3).[8\sqrt{x-1}-(3x+2)]=(x+11).[4\sqrt{x-2}-(3x-6)]+24x-8-(4x-3).(3x+2)+(x+11).(3x-6)\\\\ <=> (4x-3).[8\sqrt{x-1}-(3x+2)]=(x+11).[4\sqrt{x-2}-(3x-6)]-9x^2+52x-68\\\\ <=> (4x-3).[\frac{64(x-1)-(3x+2)^2}{8\sqrt{x-1}+(3x+2)}]=(x+11).[\frac{16(x-2)-(3x-6)^2}{4\sqrt{x-2}+(3x-6)}]-9x^2+52x-68\\\\ <=> (4x-3).\frac{-9x^2+52x-68}{8.\sqrt{x-1}+(3x+2)}=(x+11).\frac{-9x^2+52x-68}{4.\sqrt{x-2}+(3x-6)}+ (-9x^2+52x-68)\\\\ +, -9x^2+52x-68=0 <=> ....\\\\[/tex]
[tex]+, -9x^2+52x-68 \neq 0\\\\ => pt <=> \frac{4x-3}{8.\sqrt{x-1}+(3x+2)}=\frac{x+11}{4.\sqrt{x-2}+(3x-6)}+1\\\\ +, 8.\sqrt{x-1}+(3x+2)>4.\sqrt{x-2}+(3x-6)>0 (x>2)\\\\ => \frac{x+11}{4.\sqrt{x-2}+(3x-6)}>\frac{x}{4.\sqrt{x-2}+(3x-6)}>\frac{x}{8.\sqrt{x-1}+(3x+2)}\\\\ +, \frac{3x-3}{8.\sqrt{x-1}+(3x+2)} < 1 \\\\ => VT<VP => ...[/tex]
vậy....
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
