`1a)3-(5x+1)=6-4x`
`→3-5x-1=6-4x`
`→2-5x=6-4x`
`→-5x+4x=6-2`
`→-x=4`
`→x=-4`
Vậy `x=-4`
`b)(3-x)(x+5)≥0`
* `(3-x)(x+5)=0`
`→` \(\left[ \begin{array}{l}3-x=0\\x+5=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
Vậy `x∈{3;-5}`
* `(3-x)(x+5)>0`
`→` \(\left[ \begin{array}{l}3-x>0\\x+5>0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x>3\\x>-5\end{array} \right.\)
`2)`
`ax+ay+bx+by`
`=(ax+ay)+(bx+by)`
`=a(x+y)+b(x+y)`
`=(a+b)(x+y)`
Thay `a+b=-7;x+y=-18` ta có:
`(a+b)(x+y)=(-7)(-18)=126`
Vậy giá trị biểu thức là `126`
