4/(√8)
=(4*√8)/(√8*√8)=(4√8)/8=√8/2=2√2/2=√2
2) 10 - 1 + 3*x = 6*x; 10 - 1 + 3*x - 6*x = 0; 9 - 3*x = 0; -3*x = -9;откуда: x = 3.
16-88у+121у²-1=0
121у²-88у+15=0
у₁=<u>88+√(7744-7260)</u> = <u>88+22</u> = <u>110</u> = 5/11
242 242 242
у₂=<u>88-√(7744-7260)</u> = <u>88-22</u> = <u>66</u> = 3/11
242 242 242
16t²-24t+9-25t²=0
-9t²-24t+9=0
3t²+8t-3=0
t₁=<u>-8+√(64+36)</u> = <u>-8+10</u> = 1/3
6 6
t₂=<u>-8-√(64+36)</u> = <u>-8-10</u> = -3
6 6
=[ (-sina-cosa)]:{1+2sina cosa}=-(sina+cosa):(sina+cosa)^2=-1/(sina+cosa)
![\sqrt{5x^2+x} \geq 3x-1](https://tex.z-dn.net/?f=+%5Csqrt%7B5x%5E2%2Bx%7D++%5Cgeq+3x-1)
ОДЗ:
![5x^2+x \geq 0 \\ x(5x+1) \geq 0](https://tex.z-dn.net/?f=5x%5E2%2Bx+%5Cgeq+0+%5C%5C+x%285x%2B1%29+%5Cgeq+0)
a>0 ⇒ x∈(-∞;-1/5]U[0;+∞)
![5x^2+x \geq 9x^2-6x+1 \\ 4x^2-7x+1 \leq 0 \\ \\ 4x^2-7x+1=0 \\ D=49-16=33 \\ x_1= \dfrac{7- \sqrt{33} }{8} \\ x_2= \dfrac{7+ \sqrt{33} }{8}](https://tex.z-dn.net/?f=5x%5E2%2Bx+%5Cgeq+9x%5E2-6x%2B1+%5C%5C+4x%5E2-7x%2B1+%5Cleq+0+%5C%5C++%5C%5C+4x%5E2-7x%2B1%3D0+%5C%5C+D%3D49-16%3D33+%5C%5C+x_1%3D+%5Cdfrac%7B7-+%5Csqrt%7B33%7D+%7D%7B8%7D++%5C%5C+x_2%3D+%5Cdfrac%7B7%2B+%5Csqrt%7B33%7D+%7D%7B8%7D+)
Т.к.
![\sqrt{5x^2+x} = 3x-1](https://tex.z-dn.net/?f=%5Csqrt%7B5x%5E2%2Bx%7D+%3D+3x-1)
и
![\sqrt{5x^2+x} \geq 0](https://tex.z-dn.net/?f=+%5Csqrt%7B5x%5E2%2Bx%7D+%5Cgeq+0+)
то
![3x-1 \geq 0 \\ x \geq \dfrac{1}{3}](https://tex.z-dn.net/?f=3x-1+%5Cgeq+0+%5C%5C+x+%5Cgeq++%5Cdfrac%7B1%7D%7B3%7D+)
значит x=7-√33/8 - не точка смены знака в решении неравенства
С учетом ОДЗ
x∈(-∞;-1/5]U[0;7+√33/8]
Ответ: x∈(-∞;-1/5]U[0;7+√33/8]