a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

=>(3x+1)(x+2)=0

=>x=-1/3 hoặc x=-2

b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>x=-1/3 hoặc x=-4/5

9/2-1/2-1+3x=1/2+1/3

=)8/2-1+3x=5/6

=)3+3x=5/6

=)3x=5/6-3

=)3x=13/6

x=13/18

ta có: 9/2 - 1/2 - (1-3x)=1/2 + 1/3

       <=>9/2 - 1/2 - (1-3x)=5/6

       <=>1/2 - (1-3x)        =9/2-5/6

       <=>1/2 - (1-3x)        =11/3

       <=> 1 - 3x              = 19/6

       <=> 3x                   =13/6

       <=> x                     =13/18 

Tìm x ak

\(\hept{\begin{cases}3x-y=2m-1\\x+2y=3m+2\end{cases}\Leftrightarrow\hept{\begin{cases}3x-y=2m-1\\3x+6y=9m+6\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}-7y=-7m-7\\x+2y=3m+2\end{cases}}\)

\(\left(1\right)\Rightarrow y=\frac{-7\left(m+1\right)}{-7}=m+1\)(3)

Thay (3) vào (2) ta được : \(x+2m+2=3m+2\Leftrightarrow x=m\)(4)

Thay (3) ; (4) vào biểu thức trên ta được 

\(x^2+y^2=10\Rightarrow m^2+\left(m+1\right)^2=10\)

\(\Leftrightarrow m^2+m^2+2m+1=10\Leftrightarrow2m^2+2m-9=0\)

\(\Leftrightarrow m=\frac{-1\pm\sqrt{19}}{2}\)

( 3x - 6 )¹⁰ = 1

<=> ( 3x - 6 )¹⁰ = ( ±1 )¹⁰

<=> 3x - 6 = 1 hoặc 3x - 6 = -1

<=> x = 7/3 hoặc x = 5/3

\(\left(3x-6\right)^{10}=1\)   

\(\left(3x-6\right)^{10}=1^{10}\)   

\(\Rightarrow\orbr{\begin{cases}3x-6=1\\3x-6=-1\end{cases}}\)   

\(\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{5}{3}\end{cases}}\)

a) \(\Leftrightarrow\left(-63x^2+78x-15\right)+\left(63x^3+x-20\right)=44\)

\(\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\)

\(\Leftrightarrow79x-35=44\)

\(\Leftrightarrow79x=44+35\)

\(\Leftrightarrow79x=79\)

\(\Leftrightarrow x=1\)

b) \(\Leftrightarrow\left(x^2+3x+2\right).\left(x+5\right)-x^2.\left(x+8\right)=27\)

\(\Leftrightarrow x.\left(x^2+3x+2\right)+5.\left(x^2+3x+2\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x+10=27\)

\(\Leftrightarrow17x=17\)

\(\Leftrightarrow x=1\)

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

ĐK: x\(\ne\){-3;0;3}.

\(\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{3x^2}=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{3x^2}=\dfrac{1}{3x}\).