\(b,3\left(x-2\right)+2\left(3x-5\right)=10\\ \Leftrightarrow3x-6+6x-10=10\\ \Leftrightarrow3x+6x=10+10+6\\ \Leftrightarrow9x=26\\ \Leftrightarrow x=\dfrac{26}{9}\\ c,2x-\left(3x+1\right)=5x-2\\ \Leftrightarrow2x-3x-1=5x-2\\ \Leftrightarrow2x-3x-5x=-2+1\\ \Leftrightarrow-6x=-1\\ \Leftrightarrow x=\dfrac{1}{6}\\ d,3x+2=-5+6 \\ \Leftrightarrow3x=-5+6-2\\ \Leftrightarrow3x=-2\\ \Leftrightarrow x=-\dfrac{1}{3}\)

a: =>3x-6+6x-10=10

=>9x=26

=>x=26/9

b: =>5x-2=2x-3x-1

=>5x-2=-x-1

=>6x=1

=>x=1/6

d: =>3x+2=1

=>3x=-1

=>x=-1/3

a) x²-x=x(x-1)

b) 3x-3y = 3(x-y)

c)3xy²+6xyz=3xy(y+2z)

d) 5x-20y=5(x-4y)

g) 5x(x-1)-(1-x)=5x(x-1)+(x-1)=(5x+1)(x-1)

a, \(x^2-x=x\left(x-1\right)\)

b, \(3x-3y=3\left(x-y\right)\)

c, \(3xy^2+6xyz=3xy\left(y+2z\right)\)

d, \(5x-20y=5\left(x-4y\right)\)

g, \(5x\left(x-1\right)-\left(1-x\right)=\left(5x+1\right)\left(x-1\right)\)

a,\(x^2-x=x\left(x-1\right)\)

b,\(3x-3y=3\left(x-y\right)\)

c,\(3xy^2+6xyz=3xy\left(y+2z\right)\)

d,\(5x-20y=5\left(x-4y\right)\)

g,\(5x\left(x-1\right)-\left(1-x\right)=5x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(5x+1\right)\)

a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)

=-4/3x^2+8/3-10/3

=-4/3x^2-2/3

d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)

\(=3x^2+9x+22+\dfrac{68}{x-3}\)

a)

`2x-3=2-x`

`<=>2x+x=2+3`

`<=>3x=5`

`<=>x=5/3`

b)

`3x+3=7+5x`

`<=>3x-5x=7-3`

`<=>-2x=4`

`<=>x=-2`

c)

`7x-3=3x+13`

`<=>7x-3x=13+3`

`<=>4x=16`

`<=>x=4`

d)

`(5x-2)/3=(5-3x)/2`

`<=>10x-4=15-9x`

`<=>10x+9x=15+4`

`<=>19x=19`

`<=>x=1`

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

a)\(f\left(x\right)=\left(3x+4\right)\cdot\left(5x-1\right)+\left(5x+2\right)\cdot\left(1-3x\right)+2\)

\(=15x^2-3x+20x-4+5x-15x^2+2-6x+2\)

\(=16x\)

b)\(g\left(x\right)=\left(5x-1\right)\cdot\left(2x+3\right)-3\cdot\left(3x-1\right)\)

\(=10x^2+15x-2x-3-9x+3\)

\(=10x^2+4x\)

b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)

\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)

\(\Leftrightarrow3x\left(x+4\right)=0\)

=>x=0(nhận) hoặc x=-4(loại)