Question

Решить данные задания. Даю 100 баллов.

Answer 1

Ответ:

Объяснение:

1.

$\displaystyle\\\left \{ {{3(x-1)-2(1+x)<1} \atop {3x-4>0}} \right. ;\left \{ {{3x-3-2-2x<1} \atop {3x>4}} \right. ;\left \{ {{x<6} \atop {x>\dfrac{4}{3} }} \right.$

Ответ:  $\bf\\x\in\Big(1\dfrac{1}{3};6\Big)$

2.

$(\sqrt{6} +\sqrt{3} )\sqrt{12}-2\sqrt{6}\cdot\sqrt{3} =\\\\=\sqrt{6}\cdot\sqrt{12}+ \sqrt{3}\cdot\sqrt{12}-2\sqrt{6\cdot3} =\\\\=\sqrt{6\cdot12}+\sqrt{3\cdot12}-\sqrt{4\cdot18}=\\\\=\sqrt{72} +\sqrt{36} -\sqrt{72}=6$

3.

$\displaystyle\\\Big(\frac{6}{y^2-9} +\frac{1}{3-y} \Big)\cdot\frac{y^2+6y+9}{5}=\\\\\\=\Big(\frac{6}{(y-3)(y+3)} -\frac{1}{y-3} \Big)\cdot\frac{(y+3)^2}{5}=\\\\\\=\frac{6-y-3}{(y-3)(y+3)}\cdot\frac{(y+3)^2}{5}=\\\\\\=\frac{3-y}{y-3}\cdot\frac{y+3}{5}=- \frac{y+3}{5}$