`***` Lời giải chi tiết `***`

`a)`

`(x-6)(x+1)=2(x+1)`

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

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

`<=>(x+1)(x-8)=0`

`<=>` \(\left[ \begin{array}{l}x+1=0\\x-8=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-1\\x=8\end{array} \right.\) 

 Vậy phương trình có tập nghiệm là : `S={-1;8}`

`=>D`

`b)`

`(x-2)(5x+3)=(3x-8)(x-2)`

`<=>(x-2)(5x+3)-(3x-8)(x-2)=0`

`<=>(x-2)(5x+3-3x+8)=0`

`<=>(x-2)(2x+11)=0`

`<=>` \(\left[ \begin{array}{l}x-2=0\\2x+11=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=-\dfrac{11}{2}\end{array} \right.\)

Vậy phương trình có tập nghiệm là : `S={2;-(11)/(2)}`

 `=>D`

`c)`

`2x(15x+25)-33(3x+5)=0`

`<=>10x(3x+5)-33(3x+5)=0`

`<=>(3x+5)(10x-33)=0`

`<=>` \(\left[ \begin{array}{l}3x+5=0\\10x-33=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac{5}{3}\\x=\dfrac{33}{10}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là : `S={-(5)/(3);(33)/(10)}`

`d)`

`(2x^{2}+1)(2x+5)=(2x^{2}+1)(x-1)`

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

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

`<=>(2x^{2}+1)(x+6)=0`

Vì `2x^{2}+1≥1` với mọi `x` hay `2x^{2}+1\ne0` với mọi `x`

`=>x+6=0`

`<=>x=-6`

Vậy phương trình có nghiệm là : `x=-6`

`=>B`

`a)`

`(x-6)(x+1)=2(x+1)`

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

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

`<=>(x-8)(x+1)=0`

`<=>` \(\left[ \begin{array}{l}x-8=0\\x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=8\\x=-1\end{array} \right.\) 

Vậy `S={-1;8}`

`\to \text{D}`

`b)`

`(x-2)(5x+3)=(3x-8)(x-2)`

`<=>(x-2)(5x+3)-(3x-8)(x-2)=0`

`<=>(x-2)(5x+3-3x+8)=0`

`<=>(x-2)(2x+11)=0`

`<=>` \(\left[ \begin{array}{l}x-2=0\\2x+11=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=2\\x=-\dfrac{11}{2}\end{array} \right.\) 

Vậy `S={-11/2;2}`

`\to \text{D}`

`c)`

`2x(15x+25)-33(3x+5)=0`

`<=>2x.5(3x+5)-33(3x+5)=0`

`<=>10x(3x+5)-33(3x+5)=0`

`<=>(10x-33)(3x+5)=0`

`<=>` \(\left[ \begin{array}{l}10x-33=0\\3x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{33}{10}\\x=-\dfrac{5}{3}\end{array} \right.\) 

Vậy `S={5/3;33/10}`

`\to \text{D}`

`d)`

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

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

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

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

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

Vậy `S={-6}`

`\to \text{B}`