Question

1032. 1) x+1 2 x+0,5 3 1,5x12x + 1; 2) 2x+1 x+0,6 500 2409 1-1,5 x < 3,5 - x. >​

Answer 1

Ответ:

1)

Освободимся от знаменателей дробей, умножив 1 неравенство на 6 .

$\left\{\begin{array}{l}\bf \dfrac{x+1}{2}\leq \dfrac{x+0,5}{3}\ |\cdot 6\\\bf 1,5x-1\leq 2x+1\end{array}\right\ \ \left\{\begin{array}{l}\bf 3(x+1)\leq 2(x+0,5)\\\bf -1-1\leq 2x-1,5x\end{array}\right\ \ \left\{\begin{array}{l}\bf 3x+3\leq 2x+1\\\bf -2\leq 0,5x\end{array}\right$

$\left\{\begin{array}{l}\bf x\leq -2\\\bf 0,5x\geq -2\end{array}\right\ \ \left\{\begin{array}{l}\bf x\leq -2\\\bf x\geq -4\end{array}\right\ \ \ \Rightarrow \ \ \ \bf -4\leq x\leq -2\\\\\\Otvet:\ \ x\in [-4\ ;-2\ ]\ .$

2)

Освободимся от знаменателей дробей, умножив 1 неравенство на 10 .

$\left\{\begin{array}{l}\bf \dfrac{2x+1}{5}\geq \dfrac{x+0,6}{2}\ |\cdot 10\\\bf 1-1,5x < 3,5-x\end{array}\right\ \ \left\{\begin{array}{l}\bf 2(2x+1)\geq 5(x+0,6)\\\bf 1-3,5 < 1,5x-x\end{array}\right\ \ \left\{\begin{array}{l}\bf 4x+2\geq 5x+3\\\bf -2,5 < 0,5x\end{array}\right$

$\left\{\begin{array}{l}\bf -3+2\geq 5x-4x\\\bf 0,5x > -2,5\end{array}\right\ \ \left\{\begin{array}{l}\bf -1\geq x\\\bf x > -5\end{array}\right\ \ \left\{\begin{array}{l}\bf x\leq -1\\\bf x > -5\end{array}\right\ \ \ \Rightarrow \ \ \ \bf -5 < x\leq -1\\\\\\\boldsymbol{Otvet:\ \ x\in (-5\ ;-1\, ]\ .}$