-How to solve 3x ^ 2 + 2x + 1 = (- x + 1) (3x + 1)?

-How to solve 3x ^ 2 + 2x + 1 = (- x + 1) (3x + 1)?

Cross phase multiplication x = 1 x = - 1 / 3

How to solve X / (x + 1) = 2 / (3x + 3) + 4
y=x+1,x^2-2x-y=2.=

x/(x+1)=2/(3x+3)+4
Both sides of the equation multiply by 3x + 3: 3x = 2 + 4 (3x + 3) x = - 14 / 9 test: it is the solution of the original fractional equation

How to solve 2 (x-1) + 1 = 3 (3x + 1)

We get 2x-2 + 1 = 9x + 3, x = - 4 / 7

X + 3 (3x-1) = x + 3 solution

Divide both sides by X and add 3
3x-1=1
3x=2
x=2/3

How to solve 3x-4.5 * 6 = 21

3X=21+27=48,X=16

(x-1)(x-2)+3x(x+3)-4(x-2)(x+2)

(x-1)(x-2)+3x(x+3)-4(x-2)(x+2)
It unfolds well
=x^2-3x+2+3x^2+9x-4x^2+16
=6x+18

3x-4 = a (X-2) + B (x-1)

3x-4=A(x-2)+B(x-1).
3x-4=Ax+Bx-2A-B
3x-4=(A+B)x-(2A+B)
Then a + B = 3
2A+B=4
The solution is a = 1. B = 2

If x ^ 2-3x + 1 = 0, it's better to find the sum of x ^ 4 and its reciprocal instead of using the equation to find the value of X. should we use the binomial theorem?

X ^ 2-3x + 1 = 0 x ^ 2 + 1 = 3x x + 1 / x = 3, get x ^ 2 + 1 / x ^ 2 + 2 = 9 x ^ 2 + 1 / x ^ 2 = 7, get x ^ 4 + 1 / x ^ 4 + 2 = 49 x ^ 4 + 1 / x ^ 4 = 47

(3x-2) × 1 / 2 = - (x-4) process, thank you!

(3x-2) × 1 / 2 = - (x-4) both sides multiply by 2 at the same time
3x-2=-2(x-4)
3x-2=-2x+8
3x+2x=8+2
5x=10
x=2

X + 1-X-1 / 2 is greater than 4 / 3x-1

Double six on both sides
6x+6-3x+3>8x-6
3x+9>8x-6
5x