Question

Помогите пожалуйста
Буду очень благодарен

Answer 1

Объяснение:

1.

$\left \{ {{3*(x-1)-2*(1+x) < 1} \atop {3x-4 > 0}} \right.\ \ \ \ \left \{ {{3x-3-2-2x < 1} \atop {3x > 4\ |:3}} \right. \ \ \ \ \left \{ {{x < 6} \atop {x > \frac{4}{3} }} \right. \ \ \ \ \Rightarrow$

Ответ: x∈(4/3;6).

2.

$(\sqrt{6}+\sqrt{3})*\sqrt{12}-2\sqrt{6}*\sqrt{3} =\sqrt{6*12} +\sqrt{3*12} -2\sqrt{6*3}=\\ =\sqrt{72}+\sqrt{36} -2\sqrt{18} =\sqrt{36*2} +6-2\sqrt{9*2}=6\sqrt{2}+6-2*3\sqrt{2}=\\ =6\sqrt{2}+6-6\sqrt{2}=6.$

3.

$(\frac{6}{y^2-9} +\frac{1}{3-y} )*\frac{y^2+6x+9}{5} =(\frac{6}{(y+3)(y-3)}-\frac{1}{y-3})*\frac{(y+3)^2}{5} =\frac{6-1*(y+3)}{(y-3)(y+3)} *\frac{(y+3)^2}{5}=\\ =\frac{6-y-3}{(y-3)(y+3)}*\frac{(y+3)^2}{5} =\frac{3-y}{y-3}*\frac{y+3}{5}=-\frac{y-3}{y-3}*\frac{y+3}{5} =-\frac{y+3}{5} .$