Помогите решить 3 задания! 1- Найти наибольшее целое x. удовлетворяющее системе...

\frac{2-x}{2} +1 \\ \\ 0.6x-0.4+3> \frac{4x}{3} -0.5x+0.5 \\ \\ 2x-5>2-x+2 \\ 0.6x+2.6> \frac{4x}{3} -0.5x+0.5 \\ \\ 2x-5>4-x \\ 1.8x+7.8>4x-1.5x+1.5 \\ \\ 2x+x>5+4 \\ 1.8x+7.8>2.5x+1.5 \\ \\ 3x>9 \\ 2.5x-1.8x<7.8-1.5 \\ \\ x>3 \\ 0.7x<6.3 \\ \\ x>3 \\ x<9 \\ \\ x=8" alt="x-2.5> \frac{2-x}{2} +1 \\ \\ 0.6x-0.4+3> \frac{4x}{3} -0.5x+0.5 \\ \\ 2x-5>2-x+2 \\ 0.6x+2.6> \frac{4x}{3} -0.5x+0.5 \\ \\ 2x-5>4-x \\ 1.8x+7.8>4x-1.5x+1.5 \\ \\ 2x+x>5+4 \\ 1.8x+7.8>2.5x+1.5 \\ \\ 3x>9 \\ 2.5x-1.8x<7.8-1.5 \\ \\ x>3 \\ 0.7x<6.3 \\ \\ x>3 \\ x<9 \\ \\ x=8" align="absmiddle" class="latex-formula">

........3.//////////////////.9........

2- решить уравнение :
(x²2-9)-4(x²2-9)+3=0
(х²-9)(1-4)+3=0
-3(х²-9)+3=0
-3((х²-9)-1)=0
(х²-9)-1=0
х²-9-1=х²-10=0
х²=10
х=√10
х=-√10

3.
$\frac{200}{ \sqrt{5}} -2.5 \sqrt{180} -30 \sqrt{ \frac{9}{5}} = \\ \\ \frac{1}{ \sqrt{5}} (200-2.5 \sqrt{180*5} -30 \sqrt{9} )= \\ \\ \frac{1}{ \sqrt{5}} (200-2.5* \sqrt{900} -30*3)= \\ \\ \frac{1}{ \sqrt{5}} (200-2.5*30-90)= \\ \\ \frac{1}{ \sqrt{5}} (200-75-90)= \\ \\ \frac{1}{ \sqrt{5}} *35= \frac{35 \sqrt{5} }{5} =7 \sqrt{5}$