câu `1`
`a)1/(x-1)+1=x/(x-2)` `đkxđ :` `x#1;2`
`<=>(x-2)/[(x-1)(x-2)]+[(x-1)(x-2)]/[(x-1)(x-2)]=[x(x-1)]/[(x-2)(x-1)]`
`=>x-2+x^2-2x-x+2=x^2-x`
`<=>x=0(tm)`
vậy `x=0`
`b)(x+2)(x-3)=0`
`=>`\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-2(tm)\\x=3(tm)\end{array} \right.\)
vậy `x=-2` hoặc `x=3`
câu `2`
`(1-2x)/4-2<(1-5x)/8`
`=>2(1-2x)-16<1-5x`
`<=>2-4x-16<1-5x`
`<=>x<15`
vậy `x<15`
bài `3`
`a) ` ta có: `-3a+7≤-3b+7`
vì `7=7`
`=>-3a≤-3b`
vất `-3` đi
`=>a≥b`
`b)`
vì `m>n`
=>`2m>2n`
mà `1``>``-5`
`=>2m+1>2n-5`
