Х*(25-х2)=0
х=0
25-х2=0
х=0 х=5 х=-5
(-8y^6)^2-4y^3=64y^12-4y^3
![7^{-3log_{\frac{1}{7}}4}} - 5 \cdot 10^{2lg4-1}=7^{-3log_{7^{-1}}4}} - 5 \cdot 10^{lg4^{2}-lg10}= =7^{3log_{7}4}} - 5 \cdot 10^{lg \frac{16}{10}}=7^{log_{7}4^3}} - 5 \cdot 10^{lg \frac{16}{10}}= \\ =4^{3} - 5 \cdot \frac{16}{10}=64- 8 =56](https://tex.z-dn.net/?f=7%5E%7B-3log_%7B%5Cfrac%7B1%7D%7B7%7D%7D4%7D%7D+-+5+%5Ccdot+10%5E%7B2lg4-1%7D%3D7%5E%7B-3log_%7B7%5E%7B-1%7D%7D4%7D%7D+-+5+%5Ccdot+10%5E%7Blg4%5E%7B2%7D-lg10%7D%3D+%0A%0A+%3D7%5E%7B3log_%7B7%7D4%7D%7D+-+5+%5Ccdot+10%5E%7Blg+%5Cfrac%7B16%7D%7B10%7D%7D%3D7%5E%7Blog_%7B7%7D4%5E3%7D%7D++-+5+%5Ccdot+10%5E%7Blg+%5Cfrac%7B16%7D%7B10%7D%7D%3D+%5C%5C+%3D4%5E%7B3%7D+-+5+%5Ccdot+%5Cfrac%7B16%7D%7B10%7D%3D64-+8+%3D56)
логарифмы опускаем, т.к. основания равны:
![3x+0,5=x-2; \; \; 2x=-2,5; \; \; 2x=-\frac{5}{2}; \; \; x= -\frac{5}{4} =-1,25 \\ ODZ: \; x-2>0, \; \; 3x+0.5>0 \\ => \; kornei net](https://tex.z-dn.net/?f=3x%2B0%2C5%3Dx-2%3B+%5C%3B+%5C%3B+2x%3D-2%2C5%3B+%5C%3B+%5C%3B+2x%3D-%5Cfrac%7B5%7D%7B2%7D%3B+%5C%3B+%5C%3B+x%3D+-%5Cfrac%7B5%7D%7B4%7D+%3D-1%2C25+%5C%5C+ODZ%3A+%5C%3B+x-2%3E0%2C+%5C%3B+%5C%3B+3x%2B0.5%3E0+%5C%5C+%3D%3E+%5C%3B+kornei+net)
чтобы найти точки минимума и точки максимума, нужно взять производную и приравнять к нулю.
![y^{'}=\frac{4}{16} \cdot x^{3}+ \frac{3}{12} \cdot x^{2} - \frac{3 \cdot 2}{2} \cdot x= \\ =\frac{1}{4} \cdot x^{3}+ \frac{1}{4} \cdot x^{2} - 3x=\frac{1}{4}x \cdot (x^{2}+x-12)=\frac{1}{4}x \cdot ((x+4)(x-3))\\\frac{1}{4}x=0; \; \; \; x+4=0; \; \; \; x-3=0 \\x=0; \; \; \; x=-4; \; \; \; x=3](https://tex.z-dn.net/?f=y%5E%7B%27%7D%3D%5Cfrac%7B4%7D%7B16%7D+%5Ccdot+x%5E%7B3%7D%2B+%5Cfrac%7B3%7D%7B12%7D+%5Ccdot+x%5E%7B2%7D+-+%5Cfrac%7B3+%5Ccdot+2%7D%7B2%7D+%5Ccdot+x%3D+%5C%5C+%3D%5Cfrac%7B1%7D%7B4%7D+%5Ccdot+x%5E%7B3%7D%2B+%5Cfrac%7B1%7D%7B4%7D+%5Ccdot+x%5E%7B2%7D+-+3x%3D%5Cfrac%7B1%7D%7B4%7Dx+%5Ccdot+%28x%5E%7B2%7D%2Bx-12%29%3D%5Cfrac%7B1%7D%7B4%7Dx+%5Ccdot+%28%28x%2B4%29%28x-3%29%29%5C%5C%5Cfrac%7B1%7D%7B4%7Dx%3D0%3B+%5C%3B+%5C%3B+%5C%3B+x%2B4%3D0%3B+%5C%3B+%5C%3B+%5C%3B+x-3%3D0+%5C%5Cx%3D0%3B+%5C%3B+%5C%3B+%5C%3B+x%3D-4%3B+%5C%3B+%5C%3B+%5C%3B+x%3D3)
дальше фото
ответ: 0
(a^n)^m=a^(nm)
((√(3)−1)^(2))^(0.5)-((√(3)+2)^(2))^(0.5)={√(√3 - 2)^2 = |√3 - 2| = 2 - √3}=(√3-1)^(2*0.5)+<span>2 - √3=
</span>√3-1+<span>2 - √3=1</span>
1=1
вот эта формула:(a-b)^2=a^2-2ab=b^2
или есть такая: (a+b):2=a^2=2ab+b^2
аналогичная ей):
(x-1)^2=x^2-2*x*1+1^2
(х-1)^2=x^2-2x+1