Question

Решите неравенство: 1) 3-2(и-1)> 8+и; 2) 5(и+2)+14< 6-и; 3) 1/4(3+8и)больше или равно 6,25+и; 4)4(и+3)< 3(и+2); 5) 3(2и+1)> 5(и-1); 6) 3/5(5и-2/3)

Answer 1

1)
$3-2(u-1)\ \textgreater \ 8+u$
$3-2u+2\ \textgreater \ 8+u$
$5-2u\ \textgreater \ 8+u$
$-2u-u\ \textgreater \ 8-5$
$-3u\ \textgreater \ 3$
$u\ \textless \ -1$
-------------(-1)-----------------
//////////////
$u$ ∈ $(-$ ∞ $;-1)$

2)
$5(u+2)+14\ \textless \ 6-u$
$5u+10+14\ \textless \ 6-u$
$5u+24\ \textless \ 6-u$
$5u+u\ \textless \ 6-24$
$6u\ \textless \ -18$
$u\ \textless \ -3$
-------------(-3)-----------------
//////////////
$u$ ∈ $(-$ ∞ $;-3)$

3)
$\frac{1}{4} (3+8u) \geq 6,25+u$
$\frac{3}{4} +2u \geq 6 \frac{1}{4} +u$
$2u-u \geq 6 \frac{1}{4} - \frac{3}{4}$
$u \geq 5 \frac{1}{2}$
------------[5.5]-----------------
                    //////////////////
$u$ ∈ $[5.5;+$ ∞ $)$

4)
$4(u+3)\ \textless \ 3(u+2)$
$4u+12\ \textless \ 3u+6$
$4u-3u\ \textless \ 6-12$
$u\ \textless \ -6$
-------------(-6)-----------------
//////////////
$u$ ∈ $(-$ ∞ $;-6)$

5)
$3(2u+1)\ \textgreater \ 5(u-1)$
$6u+3\ \textgreater \ 5u-5$
$6u-5u\ \textgreater \ -5-3$
$u\ \textgreater \ -8$
------------(-8)-----------------
                    //////////////////
$u$ ∈ $(-8;+$ ∞ $)$

6)
$\frac{3}{5} (5u- \frac{2}{3} )\ \textless \ u+7,6$
$3u- \frac{2}{5}\ \textless \ u+7,6$
$3u- 0.4\ \textless \ u+7,6$
$3u-u\ \textless \ 7,6+0.4$
$2u\ \textless \ 8$
$u\ \textless \ 4$
-------------(4)-----------------
//////////////
$u$ ∈ $(-$ ∞ $;4)$

Answer 2

1)
3-2и+2>8+и
3и<-3
и<-1
2)
5и+10+14<6-и
6и<-18
и<-3
3)
3/4+2и>=6,25+и
и>=5,5
4)
4и+12<3и+6
и<-6
5)
6и+3>5и-5
и>-8
6)
3и-2/5<и+7,6
2и<8
и<4