Решить 3 уравнения, и 3 неравенства.Желательно подробно расписать. Буду блогодарен.

Question 1

Решить 3 уравнения, и 3 неравенства.
Желательно подробно расписать. Буду блогодарен.

Question 2

Я все задания решила ,но не могу отправить.Не проходит!

Question 3

1) $( \frac{1}{9}) ^{2x-5} = 3^{5x+10}$
$3^{-2*(2x-5)} = 3^{5x+10}$
$-2*(2x-5) = 5x+10, -4x+10 = 5x+10, x = 0$
2) $2^{x+2}*5^{x} = 400$
$2^{x}*4*5^{x} = 400$
$10^{x}= 100$
$10^{x} = 10^{2}$
$x = 2$
3) $\sqrt{2^{x}} * \sqrt{3^{x}} =36$
$\sqrt{6^{x}} =6^{2}$
$6^{ \frac{x}{2} }=6^{2}$
$\frac{x}{2}=2, x=4$
4) (0.09)^{2x}" alt="(0.3)^{x^{2}} > (0.09)^{2x}" align="absmiddle" class="latex-formula">
(0.3)^{4x}" alt="(0.3)^{x^{2}} > (0.3)^{4x}" align="absmiddle" class="latex-formula">
$x^{2}<4x$
$x^{2}-4x<0$
$x(x-4)<0$
$0<x<4$
5) $3^{ \frac{x+1}{x} } < 3^{2}$
$\frac{x+1}{x}<2$
$\frac{x+1-2x}{x}<0$
$\frac{1-x}{x}<0$
0}} \right" alt=" \left \{ {{1-x<0} \atop {x>0}} \right" align="absmiddle" class="latex-formula">
1} \atop {x>0}} \right" alt="\left \{ {{x>1} \atop {x>0}} \right" align="absmiddle" class="latex-formula">
1" alt="x>1" align="absmiddle" class="latex-formula">
0} \atop {x<0}} \right" alt=" \left \{ {{1-x>0} \atop {x<0}} \right" align="absmiddle" class="latex-formula">
$\left \{ {{x<1} \atop {x<0}} \right$
$x<0$
6) $( \frac{1}{3})^{ x^{2} -4x+3} \geq 3^{1-x}$
$3^{ -x^{2} +4x-3} \geq 3^{1-x}$
$-x^{2} +4x-3 \geq 1-x$
$-x^{2} +5x-4 \geq 0$
$x^{2} -5x+4 \leq 0$
$D=25-16=9$
$x_{1} = \frac{5-3}{2} = 1$
$x_{2} = \frac{5+3}{2} = 4$
$1\leq x\leq 4$

Question 4

$3^{-2(2x-3)}$ = $3^{5x+10}$
-4x+6=5x+10
9x=-4
x=-4/9

$2^{x+2}* 5^{x}=400$
$2^{x}*4* 5^{x}=400$
$10^{x}$ =10²
x=2

kalbim_zn · Answer 1 · 2018-03-20T21:16:04+0000

1) $( \frac{1}{9}) ^{2x-5} = 3^{5x+10}$
$3^{-2*(2x-5)} = 3^{5x+10}$
$-2*(2x-5) = 5x+10, -4x+10 = 5x+10, x = 0$
2) $2^{x+2}*5^{x} = 400$
$2^{x}*4*5^{x} = 400$
$10^{x}= 100$
$10^{x} = 10^{2}$
$x = 2$
3) $\sqrt{2^{x}} * \sqrt{3^{x}} =36$
$\sqrt{6^{x}} =6^{2}$
$6^{ \frac{x}{2} }=6^{2}$
$\frac{x}{2}=2, x=4$
4) (0.09)^{2x}" alt="(0.3)^{x^{2}} > (0.09)^{2x}" align="absmiddle" class="latex-formula">
(0.3)^{4x}" alt="(0.3)^{x^{2}} > (0.3)^{4x}" align="absmiddle" class="latex-formula">
$x^{2}<4x$
$x^{2}-4x<0$
$x(x-4)<0$
$0<x<4$
5) $3^{ \frac{x+1}{x} } < 3^{2}$
$\frac{x+1}{x}<2$
$\frac{x+1-2x}{x}<0$
$\frac{1-x}{x}<0$
0}} \right" alt=" \left \{ {{1-x<0} \atop {x>0}} \right" align="absmiddle" class="latex-formula">
1} \atop {x>0}} \right" alt="\left \{ {{x>1} \atop {x>0}} \right" align="absmiddle" class="latex-formula">
1" alt="x>1" align="absmiddle" class="latex-formula">
0} \atop {x<0}} \right" alt=" \left \{ {{1-x>0} \atop {x<0}} \right" align="absmiddle" class="latex-formula">
$\left \{ {{x<1} \atop {x<0}} \right$
$x<0$
6) $( \frac{1}{3})^{ x^{2} -4x+3} \geq 3^{1-x}$
$3^{ -x^{2} +4x-3} \geq 3^{1-x}$
$-x^{2} +4x-3 \geq 1-x$
$-x^{2} +5x-4 \geq 0$
$x^{2} -5x+4 \leq 0$
$D=25-16=9$
$x_{1} = \frac{5-3}{2} = 1$
$x_{2} = \frac{5+3}{2} = 4$
$1\leq x\leq 4$