<em>√0.25+</em> <em>√</em> 36=0.5+6=6.5
<em>√</em> 8-2*(-4)=<em>√8*8=<em>√64=8</em></em>
<span><span><span><em><em><em>√</em> 8-2*2=2</em></em></span></span></span>
<span><span><span><em><em><em>√</em> 8-2*3.5=<em><em><em>√1=1</em></em></em></em></em></span></span></span>
<span><span><span><em><em>Ответ:6.5; <em><em>8; 2; 1</em></em></em></em></span></span></span>
Х^2-4х-8х+32=(х-8)(х-4)
х(х-4)-8(х-4)=(х-8)(х-4)
(х-4)(х-8)=(х-8)(х-4) ч.т.д.
или же
х^2-12х+32=х^2-8х-4х+32
х^2-12х+32=х^2-12х+32
ч.т.д.
какой нравится такой и бери:)
Б
![\int\limits^x_0 {x(3t^2-8t+3) \, dt =t^3-4t^2+3t|^x_{0}=x^3-4x^2+3x\ \textgreater \ 0](https://tex.z-dn.net/?f=+%5Cint%5Climits%5Ex_0+%7Bx%283t%5E2-8t%2B3%29+%5C%2C+dt+%3Dt%5E3-4t%5E2%2B3t%7C%5Ex_%7B0%7D%3Dx%5E3-4x%5E2%2B3x%5C+%5Ctextgreater+%5C+0)
x(x²-4x+3)>0
x(x-1)(x-3)>0
x=0 x=1 x=3
_ + _ +
------------------(0)-----------(1)----------------(3)---------------------
x∈(0;1) U (3;∞)
![\int\limits^x_0 {t^3} \, dt =t^4/4|^x_{0}=x^4/4\ \textless \ 1/4](https://tex.z-dn.net/?f=+%5Cint%5Climits%5Ex_0+%7Bt%5E3%7D+%5C%2C+dt+%3Dt%5E4%2F4%7C%5Ex_%7B0%7D%3Dx%5E4%2F4%5C+%5Ctextless+%5C+1%2F4)
x^4-1<0
(x-1)(x+1)(t²+1)<0
x=1 x=-1
+ _ +
-----------------(-1)-----------------(1)-----------------
x∈(-1;1)
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