\(a=\sqrt{1969}+\sqrt{1971}\)

\(\Rightarrow a^2=1969+2\sqrt{1969\cdot1971}+1971\)

\(\Rightarrow a^2=2\cdot1970+2\sqrt{1969\cdot1971}\)                        (1)

\(b=2\cdot\sqrt{1970}\)

\(\Rightarrow b^2=4\cdot1970=2\cdot1970+2\cdot1970\)                   (2)

có : \(1969+1971\ge2\sqrt{1969\cdot1971}\)

\(\Rightarrow2\cdot1970\ge2\sqrt{1969\cdot1971}\)    vì 1969 khác 1971

\(\Rightarrow2\cdot1970>2\sqrt{1969\cdot1971}\)               (3)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow a^2< b^2\) mà a;b không âm

\(\Rightarrow a< b\)

Ta có : ( x² -1) (x+2)(x-3) = ( x - 1)(x² -4 )(x+5)

<=> ( x - 1 ).(x+1 )( x + 2).( x- 3) -( x -1 ). ( x - 2 )(x + 2 ) ( x + 5) =0 

<=> ( x- 1 ).( x + 2 ) .[ ( x + 1) (x - 3 )- ( x - 2) ( x + 5) ]=0 

<=> ( x -1).( x + 2) .[ (x² -2x - 3 ) - ( x² + 3x -10 )] = 0 

<=> ( x - 1 ). ( x+ 2 ) . ( - 5x + 7 ) = 0 

<=> \(\hept{\begin{cases}x-1=0\\x+2=0\\-5x+7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\\x=\frac{7}{5}\end{cases}}}\)

1) 3(x + 2) = 5x + 8

<=> 3x + 6 = 5x + 8

<=> 3x + 6 - 5x - 8 = 0

<=> -2x - 2 = 0

<=> -2x = 0 + 2

<=> -2x = 2

<=> x = -1

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

<=> 2x - 2 = 9 + x + 3

<=> 2x - 2 = 12 + x

<=> 2x - 2 - 12 - x = 0

<=> x - 14 = 0

<=> x = 0 + 14

<=> x = 14

3) 5 - (x - 6) = 4(3 - 2x)

<=> 5 - x + 6 = 12 - 8x

<=> 11 - x = 12 - 8x

<=> 11 - x - 12 + 8x = 0

<=> -1 + 7x = 0

<=> 7x = 0 + 1

<=> 7x = 1

<=> x = 1/7

\(\frac{2x-8}{6}-\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\)

\(\Leftrightarrow\frac{4\left(2x-8\right)}{24}-\frac{6\left(3x+1\right)}{24}=\frac{3\left(9x-2\right)}{24}+\frac{2\left(3x-1\right)}{24}\)

\(\Leftrightarrow\frac{8x-32}{24}-\frac{18x+6}{24}=\frac{27x-6}{24}+\frac{6x-2}{24}\)

\(\Leftrightarrow8x-32-18x-6=27x-6+6x-2\)

\(\Leftrightarrow8x-18x-27x-6x=-6-2+32+6\)

\(\Leftrightarrow-42x=30\)

\(\Leftrightarrow x=-\frac{5}{7}\)