4 года назад

-3(2.1х-1)+4.8=-6.7х+9.4

Знания

Алгебра

1 ответ:

Alina094 [58]4 года назад

0 0

-3(2,1х-1)+4,8=-6,7х+9,4
-6,3х+3+4,8=-6,7х+9,4
-6,3х+6,7х=9,4-3-4,8
0,4х=1,6
х=4

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Анна-75 [2]

$1)\; \; \frac{\sqrt{36-x^2}\cdot log_{0,5}x}{x-2} \leq 0\\\\ODZ:\; \; \left \{ {{36-x^2\geq 0\; ,} \atop {x>0\; ,\; x\ne 2}} \right. \; ,\; \left \{ {{x^2-36\leq 0\; ,} \atop {x>0\; ,\; x\ne 2}} \right. \; \left \{ {{(x-6)(x+6)\leq 0} \atop {x>0\; ,\; x\ne 2}} \right. \; \left \{ {{x\in [-6,6\, ]} \atop {x>0\; ,\; x\ne 2}} \right. \; \; \Rightarrow \\\\\underline {\; x\in (0,2)\cup (0,6\, ]\; }$

$Tak\; kak\; \; \sqrt{36-x^2}\geq 0\; ,\; to\\\\\\\left\{\begin{array}{ccc}log_{0,5}\, x\geq 0\\x-2<0\\x\in (0,2)\cup (0,6\, ]\end{array}\right\; \; ili\; \; \left\{\begin{array}{ccc}log_{0,5}\, x\leq 0\\x-2>0\\x\in (0,2)\cup (0,6\, ]\end{array}\right \\\\\\\left\{\begin{array}{ccc}x\leq 1\\x<2\\x\in (0,2)\cup (2,6\, ]\end{array}\right \; \; ili\; \; \left\{\begin{array}{ccc}x\geq 1\\x>2\\x\in (0,2)\cup (2,6\, ]\end{array}\right \\\\\\.\; \; \; \; \; x\in (0,1\, ]\qquad \qquad \quad ili\qquad \qquad x\in (2,6\, ]\\\\Otvet:\; \; x\in (0,1\, ]\cup (2,6\, ]$

$2)\; \; log_{0,2}(x-2)+log_{0,2}\, x>log_{0,2}(2x-3)\\\\ODZ:\; \; \left \{ {{x-2>0\; ,\; x>0} \atop {2x-3>0}} \right. \; \; \left \{ {{x>2\; ,\; x>0} \atop {2x>3}} \right. \; \; \left \{ {{x>2} \atop {x>1,5}} \right. \; \; \Rightarrow \; \; \underline {\; x>2\; }\\\\log_{0,2}\, (x-2)\, x>log_{0,2}(2x-3)\\\\\left \{ {{(x-2)\, x<2x-3} \atop {x>2}} \right. \; \; \left \{ {{x^2-4x+3<0} \atop {x>2}} \right. \; \; \left \{ {{(x-1)(x-3)<0} \atop {x>2}} \right. \; \; \left \{ {{x\in (1,3)} \atop {x>2}} \right.\; \; \Rightarrow \\\\Otvet:\; \; x\in (2,3)\; .$