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)
Замена у=log√₂ x
3y²-10y+3=0
D=100-36=64
y₁ = <u>10-8 </u>=2/6 =1/3
6
y₂ = <u>10+8</u> =3
6
При у=1/3
log√₂ x = 1/3
x=(√2)^(¹/₃) = ⁶√2
При у=3
log√₂ x =3
x=(√2)³ =2√2
Ответ: 1)
№1. 2cd4-3cd4+7cd3=7cd3-cd4=cd3(7-cd)
№3 2ab3-7ab3+5ab3=0