(√7+√7)-4√7=2√7-4√7=√7(2-4)=-2√7
или можно не выносить за скобки корень из 7 а сразу написать -2√7
(5х-2у)*(25х^2+10ху+4у^2)
у^3*(у-1)+(у-1)=(у^3+1)*(у-1)
0,2х-0,2-2+0,8х-2,3
1х-4,5
________
х=0,3; 0,3-4,5= -4,2
ОДЗ :
1) 16 - 16x > 0
- 16x > - 16
x < 1
2) x² - 3x + 2 > 0
(x - 1)(x - 2) > 0
+ - +
___________₀___________₀_________
1 2
/////////////////////// ///////////////////
3) x + 6 > 0
x > - 6
Окончательно : x ∈ (- 6 ; 1)
![log_{4}(16-16x)>log_{4}(x^{2}-3x+2)+log_{4}(x+6)\\\\log_{4}(16-16x)>log_{4}((x^{2}-3x+2)(x+6))\\\\16-16x>(x^{2} -3x+2)(x+6)\\\\16(1-x)-(x-1)(x-2)(x+6)>0\\\\16(1-x)+(1-x)(x-2)(x+6)>0\\\\(1-x)(16+(x-2)(x+6))>0\\\\(1-x)(16+x^{2}+4x- 12)>0\\\\(1-x)(x^{2}+4x+4)>0\\\\(x-1)(x+2)^{2}<0](https://tex.z-dn.net/?f=log_%7B4%7D%2816-16x%29%3Elog_%7B4%7D%28x%5E%7B2%7D-3x%2B2%29%2Blog_%7B4%7D%28x%2B6%29%5C%5C%5C%5Clog_%7B4%7D%2816-16x%29%3Elog_%7B4%7D%28%28x%5E%7B2%7D-3x%2B2%29%28x%2B6%29%29%5C%5C%5C%5C16-16x%3E%28x%5E%7B2%7D+-3x%2B2%29%28x%2B6%29%5C%5C%5C%5C16%281-x%29-%28x-1%29%28x-2%29%28x%2B6%29%3E0%5C%5C%5C%5C16%281-x%29%2B%281-x%29%28x-2%29%28x%2B6%29%3E0%5C%5C%5C%5C%281-x%29%2816%2B%28x-2%29%28x%2B6%29%29%3E0%5C%5C%5C%5C%281-x%29%2816%2Bx%5E%7B2%7D%2B4x-+12%29%3E0%5C%5C%5C%5C%281-x%29%28x%5E%7B2%7D%2B4x%2B4%29%3E0%5C%5C%5C%5C%28x-1%29%28x%2B2%29%5E%7B2%7D%3C0)
- - +
_________₀_____________₀___________
- 2 1
///////////////// /////////////////////////
x ∈ (- ∞ ; - 2) ∪ ( - 2 ; 1)
Окончательный ответ с учётом ОДЗ :
x ∈ (- 6 ; - 2) ∪ (- 2 ; 1)