Áp dụng dãy tỉ số bằng nhau ta có :

`(a+b-c)/c =(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(a+b+c)=(a+b+c)/(a+b+c)=1`

Ta có:

`(a+b-c)/c=1=>a+b-c=c =>a+b=2c`

`(b+c-a)/a=1=>b+c-a=a=>b+c=2a`

`(c+a-b)/b=1=>c+a-b=b=>c+a=2b`

Ta có:

`P=(1+a/b)(1+a/c)(1+c/b)`

`=(b+a)/b .(c+a)/c . (b+c)/b`

`=( (b+a)(c+a)(b+c))/(abc)`

`=(2a.2b.2c)/(abc)`

`=2.2.2`

`=8`

`x^2+5x+4`

`=x^2+4x+x+4`

`=(x+1)(x+4)`

`x^2+2x-8`

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

`=(x+4)(x-2)`

Ta có:

`(2x+5)/(x^2+5x+4)-(2x+2)/(x^2+2x-8)=3/2`

`<=> (2x+5)/((x+1)(x+4)) -(2x+2)/((x-2)(x+4))=3/2`

`<=>((2x+5)(x-2)-(2x+2)(x+1))/((x-2)(x+1)(x+4))=3/2`

`<=>(2x^2+5x-4x-10-2x^2-4x-2)/((x-2)(x+1)(x+4))=3/2`

`<=>(-3x-12)/((x-2)(x+1)(x+4))=3/2`

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

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

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

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

`<=>[(x=0(tm)),(x=2(ktm)):}`

Ta có:

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

Đặt `a=(x+1)/(x-2);b=(x-2)/(x-4) (a,b ne0)`

`=>a.b=(x+1)/(x-4)`

`=>`Phương trình trở thành:

`a^2+ab-12b^2=0`

`<=>a^2+4ab-3ab-12b^2=0`

`<=>(a+4b)(a-3b)=0`

`<=>[(a=-4b),(a=3b):}`

Th1:

`a=-4b`

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

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

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

`<=>5x^2-19x+12=0`

`<=>[(x=3),(x=4/5):}(tm)`

th2:

`a=3b`

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

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

`<=>x^2-3x-4=3x^2-12x+12`

`<=>2x^2-9x+16=0`(vô lý)

Vậy `x in {4/5;3}`

`a) ĐKXĐ: xne+-2`

`(x^2+4x+4)/(x^2-4)`

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

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

`b) (2x^2-x)/(x-1)+(x+1)/(1-x)+(2-x^2)/(x-1)(xne1)`

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

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

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

`=x-1`

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

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

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