Đáp án+Giải thích các bước giải:

`b)`

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

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

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

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

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

`<=>`$\left[\begin{matrix} x-2=0\\ 3x-1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=2\\ 3x=1\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=2\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `S={2;1/3}`

`c)`

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

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

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

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

`<=>(x+3)(3x+7)=0`

`<=>`$\left[\begin{matrix} x+3=0\\ 3x+7=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-3\\ 3x=-7\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-3\\ x=\dfrac{-7}{3}\end{matrix}\right.$

Vậy `S={-3;(-7)/3}`

`e)`

`2x^3+6x^2=x^2+3x`

`<=>2x^3+6x^2-x^2-3x=0`

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

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

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

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

`<=>`$\left[\begin{matrix} x=0\\x+3=0\\ 2x-1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=0\\x=-3\\ 2x=1\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=0\\x=-3\\ x=\dfrac{1}{2}\end{matrix}\right.$

Vậy `S={0;-3;1/2}`

Bài `11:`

`a)`

`1/(x+1)-5/(x-2)=15/((x+1)(2-x))` `(đkxđ:xne-1;xne2)`

`<=>(1(x-2)-5(x+1))/((x+1)(x-2))=(-15)/((x+1)(x-2))`

`<=>1(x-2)-5(x+1)=-15`

`<=>x-2-5x-5=-15`

`<=>x-5x=-15+2+5`

`<=>-4x=-8`

`<=>x=2` (không thỏa mãn `đkxđ`)

Vậy phương trình vô nghiệm.

`b)`

`(x-1)/(x+2)-x/(x-2)=(5x-2)/(4-x^2)` `(đkxđ:xne+-2)`

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

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

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

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

`<=>0x=0` (thỏa mãn với mọi `x∈R`)

Vậy phương trình có vô số nghiệm.

`c)`

`(x+5)/(x^2-5x)-(x-5)/(2x^2+10x)=(x+25)/(2x^2-50)` `(xne0;xne+-5)`

`<=>(x+5)/(x(x-5))-(x-5)/(2x(x+5))=(x+25)/(2(x-5)(x+5))`

`<=>((x+5).2(x+5)-(x-5).(x-5))/(2x(x+5)(x-5))=((x+25).x)/(2x(x-5)(x+5))`

`<=>(x+5).2(x+5)-(x-5).(x-5)=(x+25).x`

`<=>2x^2+20x+50-x^2+10x-25=x^2+25x`

`<=>2x^2+20x-x^2+10x-x^2-25x=-50+25`

`<=>5x=-25`

`<=>x=-5` (không thỏa mãn `đkxđ`)

Vậy phương trình vô nghiệm.

`d)`

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

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

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

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

`<=>x^2+x+1-3x^2=2x^2-2x`

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

`<=>-4x^2+3x+1=0`

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

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

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

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

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

`<=>`$\left[\begin{matrix} x-1=0\\ 4x+1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=1\\ 4x=-1\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=1\\x=\dfrac{-1}{4}\end{matrix}\right.$

Vậy `S={1;(-1)/4}`

`e)`

`7/(8x)+(5-x)/(4x^2-8x)=(x-1)/(2x(x-2))+1/(8x-16)` `(đkxđ:xne0;xne2)`

`<=>7/(8x)+(5-x)/(4x(x-2))=(x-1)/(2x(x-2))+1/(8(x-2))`

`<=>(7(x-2)+(5-x).2)/(8x(x-2))=((x-1).4+1.x)/(8x(x-2))`

`<=>7(x-2)+(5-x).2=(x-1).4+1.x`

`<=>7x-14+10-2x=4x-4+x`

`<=>7x-2x-4x-x=-4+14-10`

`<=>0x=0` (thỏa mãn với mọi `x∈R`)

Vậy phương trình có vô số nghiệm.