(x^2+2x-35)/(25-x^2)=((x+7)(x-5))/((5-x)(5+x))=-(x+7)/(5+x)
<span>1) x(lg5-1)=lg(2^x+1)-lg6
xlg5-x=lg(2^x/6+1/6)
lg5^x-lg10^x=lg(2^x/6+1/6)
lg(5^x/10^x)=lg(2^x/6+1/6)
5^x/5^x*2^x=2^x/6+1/6 5^x</span>≠0 сократим и умножим на 6
6/2^x=2^x+1
2^x=a a>0
6/a=a+1
6=a^2+a
a^2+a-6=0
a1=-3 искл посторонний корень a2=2
2^x=2
x=1
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4(1- y^2)(y^2+1)-(y^2-2)^2+( y^2+ y-1)(4- y^2)=
4y^2+4-4y^4-4y^2-y^4-4+4y^2-y+4y-y^3-4-y^2=
3y^2-6y^4-4+3^y-y^3
![-3,25\leq\frac{1-4x}{3}<1,25\\\\-3,25*3\leq\frac{1-4x}{3}*3<1,25*3\\\\-9,75\leq1-4x<3,75\\\\-9,75-1\leq1-4x-1<3,75-1\\\\-10,75\leq -4x<2,75|:-4\\\\-0,6875<x\leq2,6875\\\\x\in(-0,6875;2,6875]](https://tex.z-dn.net/?f=-3%2C25%5Cleq%5Cfrac%7B1-4x%7D%7B3%7D%3C1%2C25%5C%5C%5C%5C-3%2C25%2A3%5Cleq%5Cfrac%7B1-4x%7D%7B3%7D%2A3%3C1%2C25%2A3%5C%5C%5C%5C-9%2C75%5Cleq1-4x%3C3%2C75%5C%5C%5C%5C-9%2C75-1%5Cleq1-4x-1%3C3%2C75-1%5C%5C%5C%5C-10%2C75%5Cleq+-4x%3C2%2C75%7C%3A-4%5C%5C%5C%5C-0%2C6875%3Cx%5Cleq2%2C6875%5C%5C%5C%5Cx%5Cin%28-0%2C6875%3B2%2C6875%5D)
Ответ : 3 целых числа : 0 ; 1 ; 2
Решение смотри на фотографии