Bài 1:
`a) x^2+7x+6`
`=(x^2+x)+(6x+6)`
`=x(x+1)+6(x+1)`
`=(x+1)(x+6)`
`b) x^4+2008x^2+2007x+2008`
`=(x^4-x)+(2008x^2+2008x+2008)`
`=x(x^3-1)+2008(x^2+x+1)`
`=x(x-1)(x^2+x+1)+2008(x^2+x+1)`
`=(x^2+x+1)[x(x-1)+2008]`
`=(x^2+x+1)(x^2-x+2008)`
Bài 2:
`a) x^2-3x+2+|x-1|=0` (1)
Với `x>=1` thì
`(1) <=> x^2-3x+2+x-1=0`
`<=> x^2-2x+1=0`
`<=> (x-1)^2=0`
`<=> x-1=0`
`<=> x=1(TM)`
Với `x<1` thì
`(1)<=> x^2-3x+2+1-x=0`
`<=> x^2-4x+3=0`
`<=> x^2-x-3x+3=0`
`<=> (x-1)(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=1(l)\\x=3(l)\end{array} \right.\)
Vậy `x=1`
`b) 8(x+1/x)^2+4(x^2+1/x^2)^2-4(x^2+1/x^2)(x+1/x)^2=(x+4)^2` (2)
ĐKXĐ: `x\ne0`
`(2) <=> 8(x+1/x)^2+4[(x+1/x^2+2)-2]^2-4.[(x^2+1/x^2+2)-2](x+1/x)^2=(x+4)^2`
`<=> 8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2`
`<=> 8(x+1/x)^2+4[(x+1/x)^4-4(x+1/x)^2+4]-4[(x+1/x)^4-2(x+1/x)^2]=(x+4)^2`
`<=> 8(x+1/x)^2+4(x+1/x)^4-16(x+1/x)^2+16-4(x+1/x)^4+8(x+1/x)^2=(x+4)^2`
`<=> (x+4)^2=16`
`<=>`\(\left[ \begin{array}{l}x+4=4\\x+4=-4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0(l)\\x=-8(TM)\end{array} \right.\)
Vậy `x=-8`
