Question

решите пошу, срасибо

Answer 1

1. a)$\left \{ {{3x-2\ \textless \ 2+5x} \atop {8x\ \textgreater \ 15-2x}} \right.$
$\left \{ {{3x-5x\ \textless \ 2+2} \atop {8x+2x\ \textgreater \ 15}} \right.$
$\left \{ {{-2x\ \textless \ 4} \atop {10x\ \textgreater \ 15}} \right.$
$\left \{ {{x\ \textgreater \ -2} \atop {x\ \textgreater \ 1,5}} \right.$
x∈(1,5;+∞)
б)$\left \{ {{5x\ \textless \ 4+10x} \atop {6x+1\ \textgreater \ 1+4x}} \right.$
$\left \{ {{5x-10x\ \textless \ 4} \atop {6x-4x\ \textgreater \ 1-1}} \right.$
$\left \{ {{-5x\ \textless \ 4} \atop {2x\ \textgreater \ 0} \right.$
$\left \{ {{x\ \textgreater \ -0,8} \atop {x\ \textgreater \ 0}} \right.$
x∈(0;+∞)
2.а)$\frac{y-xy}{3} * \frac{6}{1- x^{2} }- \frac{y}{1+x} = \frac{y(1-x)*6}{(1-x)(1+x)*3}- \frac{y}{1+x} = \frac{2y}{1+x} - \frac{y}{1+x} = \frac{2y-y}{1+x}= \frac{y}{1+x}$
б)$\frac{3a}{1+c}- \frac{4}{1-c^2} * \frac{a-ac}{2}= \frac{3a}{1+c}- \frac{4*a(1-c)}{2(1-c)(1+c)} = \frac{3a}{1+c} - \frac{2a}{1+c} = \frac{3a-2a}{1+c}= \frac{a}{1+c}$
3.а)$\frac{6^{-4}*6^{-9}}{6^{-12}} = \frac{6^{-4+(-9)}}{6^{-12}}= \frac{6^{-13}}{6^{-12}} = 6^{-13-(-12)}=6^{-13+12}=6^{-1}= \frac{1}{6}$
б)$\frac{7^{-7}*7^{-8}}{7^{-13}} = \frac{7^{-7+(-8)}}{7^{-13}} = \frac{7^{-15}}{7^{-13}} =7^{-15-(-13)}=7^{-15+13}=7^{-2}= \frac{1}{7^2} = \frac{1}{49}$
4.а)$\left \{ {{x+y=-2} \atop {y^2-3x=6}} \right.$
Домножим на 3 первое уравнение системы и сложим со вторым:
$\left \{ {{3x+3y=-6} \atop {y^2-3x=6}} \right.$
$\left \{ {{x+y=-2} \atop {y^2+3y=0}} \right.$
$\left \{ {{x+y=-2} \atop {y(y+3)=0}} \right.$
$\left \{ {{x=-2-y} \atop {y_1=0;y_2=-3}} \right.$
$\left \{ {{x_1=-2-0; x_2=-2-(-3)} \atop {y_1=0;y_2=-3}} \right.$
$\left \{ {{x_1=-2; x_2=1} \atop {y_1=0;y_2=-3}} \right.$
б)$\left \{ {{x+y=5} \atop {x^2-3y=-15}} \right.$
Домножим на 3 первое уравнение системы и сложим со вторым:
$\left \{ {{3x+3y=15} \atop { x^{2} -3y=-15}} \right.$
$\left \{ {{x+y=3} \atop { x^{2} +3x=0}} \right.$
$\left \{ {{y=5-x} \atop {x(x+3)=0}} \right.$
$\left \{ {{y=5-x} \atop {x_1=0; x_2=-3}} \right.$
$\left \{ {{y_1=5-0; y_2=5-(-3)} \atop {x_1=0; x_2=-3}} \right.$
$\left \{ {{y_1=5; y_2=8} \atop {x_1=0; x_2=-3}} \right.$
5.а)$-4\ \textless \ 2x-1\ \textless \ 2$
$-4+1\ \textless \ 2x\ \textless \ 2+1$
$-3\ \textless \ 2x\ \textless \ 3$
$\frac{-3}{2} \ \textless \ x\ \textless \ \frac{3}{2}$
$-1,5\ \textless \ x\ \textless \ 1,5$
x∈(-1,5; 1,5)
б)$-6\ \textless \ 5x-1\ \textless \ 5$
$-6+1\ \textless \ 5x\ \textless \ 5+1$
$-5\ \textless \ 5x\ \textless \ 6$
$\frac{-5}{5}\ \textless \ x\ \textless \ \frac{6}{5}$
$-1\ \textless \ x\ \textless \ 1,2$
x∈(-1;1,2)
6.а)$3 x^{2} +9=12x- x^{2}$
$3 x^{2} + x^{2} -12x+9=0$
$4 x^{2} -2*2x*3+3^2=0$
$(2x-3)^2=0$
$2x-3=0$
$x= \frac{3}{2}$
$x=1,5$
б)$5 x^{2} +1=6x-4 x^{2}$
$5 x^{2} +4 x^{2} -6x+1=0$
$(3x)^2-2*3x*1+1^2=0$
$(3x-1)^2=0$
$3x-1=0$
$x= \frac{1}{3}$