`bb1.`

`-` ĐKXĐ: `{(xne4),(xne2):}`

`2/((x-4)(x-2))+(x+3)/(x-4)=(x-1)/(2-x)`

`=>2/((x-4)(x-2))+(x+3)/(x-4)=-(x-1)/(x-2)`

`=>2/((x-4)(x-2))+((x+3)(x-2))/((x-4)(x-2))=(-(x-1)(x-4))/((x-4)(x-2))`

`=>2+(x+3)(x-2)=-(x-1)(x-4)`

`=>2+x^2+x-6=-(x^2-5x+4)`

`=>x^2+x-4=-x^2+5x-4`

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

`=>2x(x-2)=0`

`=>[(x_1=0\ (tm)),(x_2=2\ (loại)):}`

Vậy nghiệm của phương trình là `x=0`

`bb2.`

ĐKXĐ: `{(xne-2),(xne-3):}`

`7/(x+2)+2/(x+3)=1/(x^2+5x+6)`

`=>7/(x+2)+2/(x+3)=1/((x+2)(x+3))`

`=>(7(x+3))/((x+2)(x+3))+(2(x+2))/((x+2)(x+3))=1/((x+2)(x+3))`

`=>7(x+3)+2(x+2)=1`

`=>7x+21+2x+4=1`

`=>9x+25=1`

`=>9x=-24`

`=>x=-8/3(tm)`

Vậy nghiệm của phương trình là `x=-8/3`

`bb3.`

ĐKXĐ: `xne+-1`

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

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

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

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

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

`=>4x=4`

`=>x=1` (ktm)

Vậy phương trình vô nghiệm

`bb4.`

ĐKXĐ: `xne+-2`

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

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

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

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

`=>x^2-x-2+5x+10=4+x^2-4`

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

`=>4x=-8`

`=>x=-2` (ktm)

Vậy phương trình vô nghiệm

`bb5.`

ĐKXĐ: `xne3,xne-1`

`x/(2x-6)+x/(2x+2)=-2x/((3-x)(x+1))`

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

`=>(x(x+1))/(2(x-3)(x+1))+(x(x-3))/(2(x-3)(x+1))=(4x)/(2(x-3)(x+1))`

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

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

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

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

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

`=>[(x_1=0\ (tm)),(x_2=3\ (loại)):}`

Vậy `x=0`

`bb6.`

ĐKXĐ: `xne0,xne-1`

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

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

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

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

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

`=>x^2-x=0`

`=>x(x-1)=0`

`=>[(x_1=0\ (loại)),(x_2=1\ (tmđk)):}`

Vậy `x=1`

`bb7.`

ĐKXĐ: `xne+-1`

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

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

`=>2/(2(x-1)(x+1))+(4(x+1))/(2(x-1)(x+1))=(3(x-1))/(2(x-1)(x+1))`

`=>2+4(x+1)=3(x-1)`

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

`=>4x+6=3x-3`

`=>x=-9`(tm)

Vậy `x=-9`

`bb8.`

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

ĐKXĐ: `xne1`

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

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

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

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

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

`<=>-x+1=0`

`<=>x=1(ktm)`

Vậy phương trình vô nghiệm

`bb9.`

ĐKXĐ: `xne2`

`(9x^2)/(x^3-8)+6/(x^2+2x+4)=3/(x-2)`

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

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

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

`=>9x^2+6x-12=3x^2+6x+12`

`=>6x^2=24`

`=>x^2=4`

`=>[(x_1=2\ (loại)),(x_2=-2\ (tm)):}`

Vậy `x=-2`

`bb10.`

ĐKXĐ: `xne-2`

`1/(x+2)-(2x-9)/(x^3+8)=2/(x^2-2x+4)`

`=>1/(x+2)-(2x-9)/((x+2)(x^2-2x+4))=2/(x^2-2x+4)`

`=>(x^2-2x+4)/((x+2)(x^2-2x+4))-(2x-9)/((x+2)(x^2-2x+4))=(2(x+2))/((x+2)(x^2-2x+4))`

`=>(x^2-2x+4)-(2x-9)=2(x+2)`

`=>x^2-2x+4-2x+9=2x+4`

`=>x^2-4x+13=2x+4`

`=>x^2-6x+9=0`

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

`=>x-3=0`

`=>x=3`(tm)

Vậy `x=3`

$\color{#0B6623}{♡^♡}\color{#0F7A2F}{𝕲}\color{#138F3C}{𝖎}\color{#17A34A}{𝖆}\color{#2EBF5E}{𝖆} \ \color{#4DD17A}{ } \ 
\color{#6FE095}{𝕻}\color{#8BE8AA}{𝖍}\color{#A6F0BF}{𝖔}
\color{#C0F7D3}{𝖓}\color{#D8FBE4}{𝖌}\color{#ECFFF1}{𝖌}\color{#0B6623}{♡^♡}$