Возводим обе части в квадрат, но:
![(x+6)(x-3) \geq 0](https://tex.z-dn.net/?f=%28x%2B6%29%28x-3%29+%5Cgeq+0)
![4(x+6)(x-3)=(3x-17)^2 \\4(x^2-3x+6x-18)=9x^2-102x+289 \\4x^2+12x-72=9x^2-102x+289 \\5x^2-114x+361=0 \\D=114^2-4*5*361=12996-7220=5776=76^2 \\x_1= \frac{114+76}{10}= 19 \\x_2= \frac{114-76}{10}= \frac{38}{10}=3,8](https://tex.z-dn.net/?f=4%28x%2B6%29%28x-3%29%3D%283x-17%29%5E2%0A%5C%5C4%28x%5E2-3x%2B6x-18%29%3D9x%5E2-102x%2B289%0A%5C%5C4x%5E2%2B12x-72%3D9x%5E2-102x%2B289%0A%5C%5C5x%5E2-114x%2B361%3D0%0A%5C%5CD%3D114%5E2-4%2A5%2A361%3D12996-7220%3D5776%3D76%5E2%0A%5C%5Cx_1%3D+%5Cfrac%7B114%2B76%7D%7B10%7D%3D+19%0A%5C%5Cx_2%3D+%5Cfrac%7B114-76%7D%7B10%7D%3D++%5Cfrac%7B38%7D%7B10%7D%3D3%2C8+)
проверяем:
![(19+6)(19-3) \geq 0 \\(3,8+6)(3,8-3) \geq 0](https://tex.z-dn.net/?f=%2819%2B6%29%2819-3%29+%5Cgeq+0%0A%5C%5C%283%2C8%2B6%29%283%2C8-3%29+%5Cgeq+0)
верно, значит уравнение имеет 2 корня
Ответ:
![x_1=19;\ x_2=3,8](https://tex.z-dn.net/?f=x_1%3D19%3B%5C+x_2%3D3%2C8)
1+4x<17
4x<17-1
4x<16
x<16:4
x<4
2x-1>4x+1
2x-4x>1+1
-2x>2
x<-1
4(x+1)-5x<3
4x+4-5x<3
4x-5x<3-4
-x<-1
x<1
Раскройте скобки и приведите подобные слагаемые:
3(-b+9) + 2,5(4-2b);
-3b+27+10-5b;
-8b+37
_________________
_________________
_________________
__________________\
<span>а2+4в2+2а+4в+2+4ав=4а+8в2+2+4ав=4а+16в+2+4ав=2(2а+8в+1+2ав)
</span>