Question

Решите показательное неравенство (Подробно)
4.6 1) 4^x<=8^sqrt(x+1) Ответ: [-1;3]
6.8 sqrt(2^x-3)>=3-2^0.5x Ответ: [2; + бесконечность)

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Answer 1

4.6

$2^{2x}\leq 2^{3\sqrt{x+1} }\\3\sqrt{x+1} \geq 2x\\\left[\begin{array}{ccc}\left \{ {{2x/3<0} \atop {x+1\geq 0}} \right. \\\left \{ {{2x/3\geq 0} \atop {x+1\geq 4x^2/9}} \right. \\\end{array}$

$\left[\begin{array}{ccc}\left \{ {{x<0} \atop {x\geq -1}} \right. \\\left \{ {{x\geq 0} \atop { 4x^2-9x-9\leq 0}} \right. \\\end{array}$

D=9*9+16*9=225

$(x-(9-15)/8)(x-(9+15)/8)\leq 0$

$\left[\begin{array}{ccc}\left \{ {{x<0} \atop {x\geq -1}} \right. \\\left \{ {{x\geq 0} \atop {(x+0,75)(x-3)\leq 0}} \right. \\\end{array}$

Ответ: x∈[-1;3]

6.8

$\sqrt{2^x-3} \geq 3-2^{0.5x}\\2^x=a\\\sqrt{a-3}\geq 3-\sqrt{a} \\\left[\begin{array}{ccc}\left \{ {{3-\sqrt{a}<0} \atop {a-3\geq 0}} \right. \\\left \{ {{3-\sqrt{a}\geq 0} \atop {a-3\geq 9+a-6\sqrt{a} }} \right. \\\end{array}$

$\left[\begin{array}{ccc}\left \{ {{\sqrt{a}>3} \atop {a\geq 3}} \right. \\\left \{ {\sqrt{a}\leq 3} \atop {\sqrt{a} \geq 2}} \right. \\\end{array}$

$\left[\begin{array}{ccc}\left \{ {{a>9} \atop {a\geq 3}} \right. \\\left \{ {0\leq a\leq 9} \atop {a \geq 4}} \right. \\\end{array}$

$\left \{ {{2^x>9} \atop {4\leq 2^x\leq 9}} \right. \\\left \{ {{x>log2(9)} \atop {log2(4)\leq x\leq log2(9)}} \right. \\2\leq x$

Ответ: [2;+∞)

Answer 2

Решение задания приложено 4. 6