Đáp án:

Giải thích các bước giải:

`11)(2x-3)^2=-24x`

`<=>4x^2-12x+9=-24x`

`<=>4x^2+12x+9=0`

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

`<=>2x+3=0`

`<=>x=-3/2`

Vậy `S={-3/2}`

`12)(2x+1)^2=-x`

`<=>4x^2+4x+1=-x`

`<=>4x^2+5x+1=0`

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

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

Vậy `S={-1;-1/4}`

`13)` `(x^2)/5-(2x)/3=(x+5)/6`

`<=>(6x^2-20x)/30=(5(x+5))/30`

`=>6x^2-20x=5x+25`

`<=>6x^2-25x-25=0`

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

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

Vậy `S={-5/6;5}`

`14)` `(x^2)/2-(2x)/3=(x-2)/6`

`<=>(3x^2-4x)/6=(x-2)/6`

`=>3x^2-4x=x-2`

`<=>3x^2-5x+2=0`

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

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

`15)` `(x+2)/(x-5)+3=6/(2-x)` (đk: `x\ne5;x\ne2`)

`<=>(x+2)/(x-5)+3=(-6)/(x-2)`

`<=>((x+2)(x-2)+3(x-5)(x-2))/((x-5)(x-2))=(-6(x-5))/((x-2)(x-5))`

`=>x^2-4+3x^2-21x+30=-6x+30`

`<=>4x^2-21x+26=-6x+30`

`<=>4x^2-15x-4=0`

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

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

Vậy `S={-1/4;4}`

`16)` `4/(x+1)=(-x^2-x+2)/((x+1)(x+2)` (đk: `x\ne-1;x\ne-2`)

`<=>(4(x+2))/((x+1)(x+2))=(-x^2-x+2)/((x+1)(x+2))`

`=>4x+8=-x^2-x+2`

`<=>x^2+5x+6=0`

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

`<=>` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2(ktmđk)\\x=-3(tm)\end{array} \right.\) 

Vậy `S={-3}`

$\text{Đáp án + Giải thích các bước giải:}$

`11//(2x-3)^{2}=-24x`

`<=>4x^{2}-12x+9+24x=0`

`<=>4x^{2}+12x+9=0`

`<=>x^{2}+3x+(9)/(4)=0`

`<=>(x+(3)/(2))^{2}=0`

`<=>x=-(3)/(2)`

`\text{Vậy}` `S={-(3)/(2)}`

`12//(1+2x)^{2}=-x`

`<=>4x^{2}+4x+1+x=0`

`<=>4x^{2}+5x+1=0`

`<=>(4x^{2}+4x)+(x+1)=0`

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

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

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

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

`\text{Vậy}` `S={-1;-(1)/(4)}`

`13//(x^{2})/(5)-(2x)/(3)=(x+5)/(6)`

`<=>(6x^{2})/(30)-(20x)/(30)=(5(x+5))/(30)`

`=>6x^{2}-20x=5x+25`

`<=>6x^{2}-25x-25=0`

`<=>(6x^{2}-30x)+(5x-25)=0`

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

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

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

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

`\text{Vậy}` `S={5;-(5)/(6)}`

`14//(x^{2})/(2)-(2x)/(3)=(x-2)/(6)`

`<=>(3x^{2})/(6)-(4x)/(6)=(x-2)/(6)`

`=>3x^{2}-4x=x-2`

`<=>3x^{2}-5x+2=0`

`<=>(3x^{2}-3x)-(2x-2)=0`

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

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

`<=>` \(\left[ \begin{array}{l}x-1=0\\3x-2=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=1\\x=\dfrac{2}{3}\end{array} \right.\) 

`\text{Vậy}` `S={1;(2)/(3)}`

`15//(x+2)/(x-5)+3=(6)/(2-x)` `(ĐKXĐ:x\ne{5;2})`

`<=>((x+2)(x-2))/((x-5)(x-2))+(3(x-5)(x-2))/((x-5)(x-2))=-(6(x-5))/((x-2)(x-5))`

`=>(x+2)(x-2)+3(x-5)(x-2)=-6(x-5)`

`<=>x^{2}-4+3(x^{2}-7x+10)=-6x+30`

`<=>x^{2}-4+3x^{2}-21x+30+6x-30=0`

`<=>4x^{2}-15x-4=0`

`<=>(4x^{2}-16x)+(x-4)=0`

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

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

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

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

`\text{Vậy}` `S={4;-(1)/(4)}`

`16//(4)/(x+1)=(-x^{2}-x+2)/((x+1)(x+2))` `(ĐKXĐ:x\ne{-1;-2})`

`<=>(4(x+2))/((x+1)(x+2))=(-x^{2}-x+2)/((x+1)(x+2))`

`=>4(x+2)=-x^{2}-x+2`

`<=>4x+8+x^{2}+x-2=0`

`<=>x^{2}+5x+6=0`

`<=>(x^{2}+2x)+(3x+6)=0`

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

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

`<=>` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-2(KTM)\\x=-3(TM)\end{array} \right.\) 

`\text{Vậy}` `S={-3}`