a) `7x^2 - 35x + 42 = 0`

`<=> 7(x^2 - 5x + 6) = 0`

`<=> x^2 - 5x + 6 = 0`

`<=> x^2 - 2x - 3x + 6 = 0`

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

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

`<=> x - 2 = 0` hoặc `x - 3 = 0`

`<=> x = 2` hoặc `x = 3`

Vậy `x = {2; 3}`

b) `|x+2| + |x+3| = 4(x - 1) - |x-1| `

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

Ta biết `{(|x+2|>=0),(|x+3|>=0),(|x-1|>=0):} AA x`

`<=> 4(x - 1) >= 0 <=> x - 1 >= 0 <=> x >= 1`

Từ `x >= 1` ta có: `{(|x+2| = x+2),(|x+3|=x+3),(|x-1|=x-1):}`

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

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

`<=> 3x - 4x = - 4 - 4`

`<=> -x = -8`

`<=> x = 8  (tm)`

Vậy `x = 8`

`color{cyan}{-NeverGiveUp-}`