2y^2=5 3x^2=6 9a^2=16 2x^2-25=0 x^2+8x+16=3 x^2-2x+1=4 solve equations

2y^2=5 3x^2=6 9a^2=16 2x^2-25=0 x^2+8x+16=3 x^2-2x+1=4 solve equations

2y^2=5,y^2=5/2,y=±√(5/2)=±√10/2,y1=√10/2,y2=-√10/2.3x^2=6,x^2=2,x=±√2,x1=√2,x2=-√2.9a^2=16,a^2=16/9,a=±√(16/9)=±4/3,a1=4/3,a2=-4/3.2x^2-25=0,x^2=25/2,x=±√(25/2)=±5√2/2,x1=5√2/2,x2=-5...

(2x-3)2-(1-x)2=0 9（2x-3）2=25(1-3x)2
1. (2x-3) quadratic - (1-x) quadratic = 0
2. 9 (2x-3) quadratic = 25 (1-3x) quadratic

1. (2x-3) quadratic - (1-x) quadratic = 0
（2x-3+1-x）(2x-3-1+x)=0
(x-2)(3x-4)=0
x1=2 x2=4/3
2.9 (2x-3) quadratic = 25 (1-3x) quadratic
(6x-9) quadratic = (5-15x) quadratic
(6x-9) quadratic - (5-15x) quadratic = 0
(6x-9+5-15x)(6x-9-5+15x)=0
(-9x-4)(21x-14)=0
(9x+4)(3x-2)=0
x1=-4/9 x2=2/3

Given x + y + Z = 1, XY + YZ + ZX = XYZ, prove: (1-x) (x + YZ) = 0, (1-y) (y + ZX) = 0, (1-z) (Z + XY) = 0
I'm rather dull (new, poorer, please understand)

(1-x)(x+yz)=(y+z)(x+yz)=xy+xz+yz(y+z)=xyz-yz+yz(y+z)=xyz-yz(1-yz)=xyz-xyz=0
The others are the same

Cos (π - θ) Tan (3 π + θ) / sin (π / 2 + θ) = 3 / 4, then cos θ =?

cos(π-θ)tan(3π+θ)/sin(π/2+θ)=3/4
(-cosθ)tanθ/cosθ=3/4
-tanθ=3/4
tanθ=-3/4
When θ is the second quadrant angle, cos θ = - 4 / 5
When θ is the fourth quadrant angle, cos θ = 4 / 5