`a)`

`(3x+1)^2-(2x-5)^2=0`

`⇔(3x+1-2x+5)(3x+1+2x-5)=0`

`⇔(x+6)(5x-4)=0`

`1)x+6=0⇔x=-6`

`2)5x-4=0⇔x=4/5`

Vậy `S={-6;4/5}`

`b)`

`(x+6)(3x+1)+x^2-36=0`

`⇔(x+6)(3x+1)+(x^2-6^2)=0`

`⇔(x+6)(3x+1)+(x-6)(x+6)=0`

`⇔(x+6)(3x+1+x-6)=0`

`⇔(x+6)(4x-5)=0`

`1)x+6=0⇔x=-6`

`2)4x-5=0⇔x=5/4`

Vậy `S={-6;5/4}`

`c)`

`(x+3)(4-3x)=x^2+6x+9`

`⇔(x+3)(4-3x)=(x+3)^2`

`⇔(x+3)(4-3x)-(x+3)^2=0`

`⇔(x+3)(4-3x-x-3)=0`

`⇔(x+3)(-4x+1)=0`

`1)x+3=0⇔x=-3`

`2)-4x+1=0⇔x=1/4`

Vậy `S={-3;1/4}`

`d)(4x+3)^2-4(x-1)^2=0`

`⇔(4x+3)^2-[2(x-1)]^2=0`

`⇔[4x+3-2(x-1)].[4x+3+2(x-1)]=0`

`⇔(4x+3-2x+2)(4x+3+2x-2)=0`

`⇔(2x+5)(6x+1)=0`

`1)2x+5=0⇔x=-5/2`

`2)6x+1=0⇔x=-1/6`

Vậy `S={-5/2;-1/6}`

`e)(x+5)^2(3x+2)^2=x^2(x+5)^2`

`⇔(x+5)^2(3x+2)^2-x^2(x+5)^2=0`

`⇔(x+5)^2[(3x+2)^2-x^2]=0`

`⇔(x+5)^2(3x+2-x)(3x+2+x)=0`

`⇔(x+5)^2(2x+2)(4x+2)=0`

`1)(x+5)^2=0⇔x+5=0⇔x=-5`

`2)2x+2=0⇔x=-1`

`3)4x+2=0⇔x=-1/2`

Vậy `S={-5;-1;-1/2}`

`f)(2x-1)(x-3)^2=(2x+1)(2x-1)^2`

`⇔(2x-1)(x-3)^2-(2x+1)(2x-1)^2=0`

`⇔(2x-1)[(x-3)^2-(2x+1)(2x-1)]=0`

`⇔(2x-1)(x^2-6x+9-4x^2+1)=0`

`⇔(2x-1)(-3x^2-6x+10)=0`

`1)2x-1=0⇔x=1/2`

`2)-3x^2-6x+10=0⇔x=``+-`$\dfrac{\sqrt[]{39} }{3}$`-1` 

Vậy `S={1/2;` `+-`$\dfrac{\sqrt[]{39} }{3}-1$`}`

`a)`

`(3x+1)^2-(2x-5)^2=0`

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

`<=>(x+6)(5x-4)=0`

`<=>` \(\left[ \begin{array}{l}x+6=0\\5x-4=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-6\\x=\dfrac{4}{5}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={-6;4/5}`

$\\$

`b)`

`(x+6)(3x+1)+x^2-36=0`

`<=>(x+6)(3x+1)+(x-6)(x+6)=0`

`<=>(x+6)(3x+1+x-6)=0`

`<=>(x+6)(4x-5)=0`

`<=>` \(\left[ \begin{array}{l}x+6=0\\4x-5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-6\\x=\dfrac{5}{4}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={-6;5/4}`

$\\$

`c)`

`(x+3)(4-3x)=x^2+6x+9`

`<=>(x+3)(4-3x)=(x+3)^2`

`<=>(x+3)(4-3x)-(x+3)^2=0`

`<=>(x+3)(4-3x-x-3)=0`

`<=>(x+3)(1-4x)=0`

`<=>` \(\left[ \begin{array}{l}x+3=0\\1-4x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{1}{4}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={-3;1/4}`

$\\$

`d)`

`(4x+3)^2-4(x-1)^2=0`

`<=>(4x+3)^2-[2(x-1)]^2=0`

`<=>(4x+3)^2-(2x-2)^2=0`

`<=>(4x+3-2x+2)(4x+3+2x-2)=0`

`<=>(2x+5)(6x+1)=0`

`<=>` \(\left[ \begin{array}{l}2x+5=0\\6x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-\dfrac{5}{2}\\x=-\dfrac{1}{6}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={-5/2;-1/6}`

$\\$

`e)`

`(x+5)^2(3x+2)^2=x^2(x+5)^2`

`<=>[(x+5)(3x+2)]^2=[x(x+5)]^2`

`<=>(3x^2+17x+10)^2=(x^2+5x)^2`

`<=>(3x^2+17x+10-x^2-5x)(3x^2+17x+10+x^2+5x)=0`

`<=>(2x^2+12x+10)(4x^2+22x+10)=0`

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

`<=>(x+1)(x+5)^2(2x+1)=0`

`<=>` \(\left[ \begin{array}{l}x+1=0\\(x+5)^2=0\\2x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-1\\x=-5\\x=-\dfrac{1}{2}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={-1;-5;-1/2}`

$\\$

`f)`

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

`<=>(2x-1)(x-3)^2-(2x+1)(2x-1)^2=0`

`<=>(2x-1)[(x-3)^2-(2x+1)(2x-1)]=0`

`<=>(2x-1)(x^2-6x+9-4x^2+1)=0`

`<=>(2x-1)(-3x^2-6x+10)=0`

`<=>` \(\left[ \begin{array}{l}2x-1=0\\-3x^2-6x+10=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-1-\dfrac{1}{3}\sqrt{39}\\x=-1+\dfrac{1}{3}\sqrt{39}\end{array} \right.\) 

Vậy phương trình có tập nghiệm `S={1/2;-1+-\sqrt{39}}`