Решите ,пожалуйста,очень нужно!

0 голосов
35 просмотров

Решите ,пожалуйста,очень нужно!


image

Алгебра (9.2k баллов) | 35 просмотров
Дан 1 ответ
0 голосов
Правильный ответ
2^{x+5}\leq (\frac{1}{4})^{2x-3}
\\\
2^{x+5}\leq (2^{-2})^{2x-3}
\\\
2^{x+5} \leq 2^{-4x+6}
\\\
x+5 \leq -4x+6
\\\
5x \leq 1
\\\
x \leq 0.2

0.7^x \geq 0.49^{5-x}
\\\
0.7^x \geq 0.7^{10-2x}
\\\
x \leq 10-2x
\\\
3x \leq 10
\\\
x \leq \frac{10}{3}

image64^{x^2-4} \\\ (2^{-2} )^{3x}>(2^6)^{x^2-4} \\\ 2^{-6x}>2^{6x^2-24} \\\ -6x>6x^2-24 \\\ 6x^2+6x-24<0 \\\ x^2+x-4<0 \\\ D=1+4\cdot4=17 \\\ x= \frac{-1\pm \sqrt{17} }{2} \\\ x\in(\frac{-1-\sqrt{17} }{2} ; \frac{-1+\sqrt{17} }{2} )" alt="( \frac{1}{4} )^{3x}>64^{x^2-4} \\\ (2^{-2} )^{3x}>(2^6)^{x^2-4} \\\ 2^{-6x}>2^{6x^2-24} \\\ -6x>6x^2-24 \\\ 6x^2+6x-24<0 \\\ x^2+x-4<0 \\\ D=1+4\cdot4=17 \\\ x= \frac{-1\pm \sqrt{17} }{2} \\\ x\in(\frac{-1-\sqrt{17} }{2} ; \frac{-1+\sqrt{17} }{2} )" align="absmiddle" class="latex-formula">

image0}} \right. \\\ 0 0}} \right. \\\ 0

image( \frac{3}{4})^{-2} } \atop {4x+3>0}} \right. \\\ \left \{ {{4x+3> \frac{16}{9} } \atop {x>- \frac{4}{3} }} \right. \\\ \left \{ {{4x>- \frac{11}{9} } \atop {x>- \frac{4}{3} }} \right. \\\ \left \{ {{x>- \frac{11}{36} } \atop {x>- \frac{4}{3} }} \right. \\\ x>- \frac{11}{36}" alt="\log_{ \frac{3}{4} }(4x+3)<-2 \\\ \left \{ {{4x+3>( \frac{3}{4})^{-2} } \atop {4x+3>0}} \right. \\\ \left \{ {{4x+3> \frac{16}{9} } \atop {x>- \frac{4}{3} }} \right. \\\ \left \{ {{4x>- \frac{11}{9} } \atop {x>- \frac{4}{3} }} \right. \\\ \left \{ {{x>- \frac{11}{36} } \atop {x>- \frac{4}{3} }} \right. \\\ x>- \frac{11}{36}" align="absmiddle" class="latex-formula">

image0}} \right. \\\ \left \{ {{-x \leq 8} \atop {x> 9}} \right. \\\ \left \{ {{x \geq -8} \atop {x >9}} \right. \\\ x >9" alt="\lg(x-9) \leq \lg(2x-1) \\\ \left \{ {{x-9 \leq 2x-1} \atop {x-9 >0}} \right. \\\ \left \{ {{-x \leq 8} \atop {x> 9}} \right. \\\ \left \{ {{x \geq -8} \atop {x >9}} \right. \\\ x >9" align="absmiddle" class="latex-formula">

image2 \\\ 3x+1>7^2 \\\ 3x+1>49 \\\ 3x>48 \\\ x>12" alt="\log_7(3x+1)>2 \\\ 3x+1>7^2 \\\ 3x+1>49 \\\ 3x>48 \\\ x>12" align="absmiddle" class="latex-formula">

image0}} \right. \\\ \left \{ {{2x \leq8} \atop {x> \frac{7}{3} }} \right. \\\ \left \{ {{x \leq4} \atop {x> \frac{7}{3} }} \right. \\\ \frac{7}{3} 0}} \right. \\\ \left \{ {{2x \leq8} \atop {x> \frac{7}{3} }} \right. \\\ \left \{ {{x \leq4} \atop {x> \frac{7}{3} }} \right. \\\ \frac{7}{3}

image\log_{0.5}(3-2x) \\\ \left \{ {{x<3-2x} \atop {x>0}} \right. \\\ \left \{ {{3x<3} \atop {x>0}} \right. \\\ \left \{ {{x<1} \atop {x>0}} \right. \\\ 0\log_{0.5}(3-2x) \\\ \left \{ {{x<3-2x} \atop {x>0}} \right. \\\ \left \{ {{3x<3} \atop {x>0}} \right. \\\ \left \{ {{x<1} \atop {x>0}} \right. \\\ 0
(271k баллов)
0

Спасибо Вам огромное,очень выручили,очень благодарна!!!