Question

Rezolvati in multimea nr.intregi urm.inecuatii:
a) (3x+1)²-(x-1)(x+1)-11≤2(3-2x)²+(x+1)²-x²
b) (2x+3)/2-1/5<3-(2x+5)/10+(x-1)/5

Answer 1

a)

(3x+1)²-(x-1)(x+1)-11≤2(3-2x)²+(x+1)²-x²

9x²+6x+1-x²+1-11≤18-24x+8x²+x²+2x+1-x²

8x²+6x-9≤18-24x+8x²+x²+2x+1-x²

8x²+6x-9≤19-22x+8x²

6x-9≤19-22x

6x≤19-22x+9

6x≤-22x+28

6x+22x≤28

28x≤28

x≤1

x € (-∞;1]

$b) \\ \\ \frac{(2x + 3)}{2} - \frac{1}{5} < 3 - \frac{(2x + 5)}{10} + \frac{(x - 1)}{5}$

5(2x+3)-2<30-2x-5+2(x-1)

5(2x+3)-2<25-2x+2(x-1)

10x+15-2<25-2x+2x-2

10x+13<25-2x+2x-2

10x+13<25-2

10x+13<23

10x<23-13

10x<10

x<1

x € (-∞;1)

Answer 2

$a) (3x+1)^{2}-(x-1)(x+1)-11\leq 2(3-2x)^{2}+(x+1)^{2}-x^{2} \\ 9x^{2} +6x+1-(x^{2} -1^{2})-11\leq 2(9-12x+4x^{2}) +x^{2} +2x+1-x^{2} \\ 9x^{2} +6x+1-x^{2} +1-11\leq 18-24x+8x^{2} +x^{2} +2x+1-x^{2} \\ 8x^{2} +6x-9\leq 8x^{2} -22x+19\\ 8x^{2} +6x-9-8x^{2} +22x-19\leq 0\\ 28x-28\leq 0\\28x\leq 28\\x\leq \frac{28}{28} \\x\leq 1$

x∈(-∞; 1]

$b) \frac{2x+3}{2} -\frac{1}{5} <3-\frac{2x+5}{10}+\frac{x-1}{5} \\ \frac{5(2x+3)-2}{10} <\frac{30-2x-5+2(x-1)}{10}\\ \frac{10x+15-2}{10}<\frac{30-2x-5+2x-2}{10}\\ \frac{10x+13}{10} <\frac{23}{10}\\ \frac{10x+13}{10}-\frac{23}{10}<0\\\frac{10x+13-23}{10}<0\\\frac{10x-10}{10} \\10x-10<0\\10x<10\\ x<1$

x∈(-∞;1)