Учитывая, что тангенс неопределен в точках пи-пополам, три пи-пополам и так далее, подойдут не все решения, а только
<span>(5√7-√63+√14)•√7
(5</span>√7-3√7+√14)*√7
(2√7+√14)*√7
2*7+√14*√7
14+7√2
Ответ:14+7√2
{x²- 4≥0, log₁/₇(x+2) ≤ <span>-1
</span>x²- 4≥0 ⇒ x² ≥4, -2≥ х ≥2 , х∈(-∞;-2]∪[2;+∞)
log₁/₇(x+2) ≤ -1
1/7<1 значит при решении знак меняем на противоположный
(x+2)≥(1/7)⁻¹
(x+2)≥ 7 х≥5 х∈[5; +∞)
Ответ х∈[5; +∞)
А)<span><span>(<span><span>x2</span>+x</span>)</span><span>(<span>49−<span>x2</span></span>)</span>=</span>Раскрытие скобок:<span><span>x2</span>49+<span>x4</span><span>(<span>−1</span>)</span>+x49+<span>x3</span><span>(<span>−1</span>)</span>=</span><span>49<span>x2</span>−<span>x4</span>+49x−<span>x3</span>=</span><span>−<span>x4</span>−<span>x3</span>+49<span>x2</span>+49x</span>Ответ: <span>−<span>x4</span>−<span>x3</span>+49<span>x2</span>+49x
В)</span><span><span>(<span><span>x2</span>−7</span>)</span><span>(<span><span>x2</span>+18</span>)</span>=</span>Раскрытие скобок:<span><span>x4</span>+<span>x2</span>18−7<span>x2</span>−126=</span><span><span>x4</span>+11<span>x2</span>−126</span>Ответ: <span><span>x4</span>+11<span>x2</span>−126
Д)</span><span><span>(<span><span>x2</span>−3<span>x2</span></span>)</span><span>(<span><span>x2</span>+7</span>)</span>=</span>Приведение подобных:<span><span>(<span>−2<span>x2</span></span>)</span><span>(<span><span>x2</span>+7</span>)</span>=</span>Раскрытие скобок:<span>−2<span>x4</span>−14<span>x2</span></span>Ответ: <span><span>−2<span>x4</span>−14<span>x2
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