Решите неравенство: 1) 3-2(и-1)>8+и; 2) 5(и+2)+14

1)-------------(-1)-----------------////////////// ∈ ∞ 2)-------------(-3)-----------------////////////// ∈ ∞ 3)------------[5.5]----------------- ////////////////// ∈ ∞ 4)-------------(-6)-----------------////////////// ∈ ∞ 5)------------(-8)----------------- ////////////////// ∈ ∞ 6)-------------(4)-----------------////////////// ∈ ∞

3)

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Verified answer

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)-----------------
//////////////
$u$ ∈ $(-$ ∞ $;4)$

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sdelajte pozhalujsta 1 i 2 tapsyrma srochno

3 pzhzh ochen nado

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