2(x - 3) + 5 = 3x - 1

2x-6+5=3x-1

2x-1=3x-1

2x-3x=-1+1

-x=0

x=0

2x(3x + 2) - 5 = 3( 2x^2 - 2x + 1)

6x²+4x-5=6x²-6x+3

6x²+4x-6x²+6x=3+5

10x=8

x=4/5

(3x - 2)(2x - 3) + 5 = 5

(3x-2)(2x-3)=0

=>3x-2=0 hoặc 2x-3=0

=>x=2/3 hoặc x=3/2

2(x - 3) + 5 = 3x - 1

<=>2x-6+5=3x-1

<=>2x-3x=-1+6-5

<=>-x=0

<=>x=0

2x(3x + 2) - 5 = 3( 2x² - 2x + 1)

<=>6x²+4x-5=6x²-6x+3

<=>4x+6x=3+5

<=>10x=8

<=>x=0,8

(3x - 2)(2x - 3) + 5 = 5

<=>(3x-2)(2x-3)=0

<=>3x-2=0 hoặc 2x-3=0

<=>x=2/3 hoặc x=3/2

2x^3+6x^2=x^2+3x

=>2x^3+6x^2-x^2-3x=0

=>2x^3+5x^2-3x=0

=>x(2x^2+5x-3)=0

=>x=0 hoặc 2x^2+5x-3=0

=>x=0  ,       2x^2-x+6x-3=0

=>x=0  ,        x(2x-1)+3(2x-1)=0

=>x=0  ,        (2x-1)(x+3)=0

=>x=0  ,   2x-1=0 hoac x+3=0

=> x=0 ,   x=1/2   ,x=-3

giải ra giùm cái đy chi tiết vô

1: 6(x-2)-y(2-x)=10

=>6(x-2)+y(x-2)=10

=>(x-2)(y+6)=10

=>(x-2;y+6)∈{(1;10);(10;1);(-1;-10);(-10;-1);(2;5);(5;2);(-2;-5);(-5;-2)}

=>(x;y)∈{(3;4);(12;-5);(1;-16);(-8;-7);(4;-1);(7;-4);(0;-11);(-3;-8)}

2: 3x-2xy+3y=6

=>x(3-2y)+3y-4,5=6-4,5

=>-x(2y-3)+1,5(2y-3)=1,5

=>(2y-3)(-x+1,5)=1,5

=>(2y-3)(-2x+3)=3

=>(2x-3)(2y-3)=-3

=>(2x-3;2y-3)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(2;0);(0;2);(1;3);(3;1)}

3: 6x-xy+2y=5

=>x(6-y)+2y-12=5-12=-7

=>-x(y-6)+2(y-6)=-7

=>(y-6)(-x+2)=-7

=>(x-2)(y-6)=7

=>(x-2;y-6)∈{(1;7);(7;1);(-1;-7);(-7;-1)}

=>(x;y)∈{(3;13);(9;7);(1;-1);(-5;5)}

(2x - 1)² - 3/2 = 3x

<=> 4x² - 4x + 1 - 3/2 - 3x = 0

<=> 4x² - 7x - 1/2 = 0

\(=>x = {7 \pm \sqrt{57} \over 8}\)

\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)

a: \(=\dfrac{\left(3x-1\right)^3}{3x-1}\cdot\dfrac{8xy}{12x^3}=\dfrac{2y\left(3x-1\right)^2}{3x^2}\)

b: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{x\left(x-5\right)}=\dfrac{x+5}{x}\)