Bài 1:
a) `-1/12 - ( 2 5/8 - 1/3)`
`= -1/12 - (21/8 - 1/3)`
`= -1/12 -21/8 + 1/3`
`=-65/24 + 1/3`
`=-19/8`
b) `2/5 + (-4/3) + (-1/2)`
`= -14/15 + (-1/2)`
`=-43/30`
c) `3/12 - (6/15 -3/10)`
`= 1/4 - 2/5+ 3/10`
`=-3/20 + 3/10`
`=3/20`
Bài 2:
a) `|x-1| = 2x-5`
Với mọi `x` ta luôn có: `|x-1| ge 0 => 2x-5 ge 0=> 2x ge 5 => x ge 5/2`
+) `x-1 = 2x-5`
`=> x - 2x = -5 +1`
`=> -x = -4`
`=> x= 4 ` (nhận)
+) `x -1 = -2x +5`
`=> x + 2x = 5 +1`
`=> 3x=6`
`=> x= 6:3`
`=> x= 2` (loại do `2 < 5/2`)
Vậy `x=4`
b) `| 9 -7x| = 5x-3`
Với mọi `x` ta luôn có: `| 9- 7x| ge 0 => 5x-3 ge 0 => 5x ge 3 => x ge 3/5`
+) `9-7x = 5x -3`
`=> 9 + 3 = 5x + 7x`
`=> 12 = 12x`
`=>x = 1` (nhận)
+) `9-7x = -5x + 3`
`=> 9 -3 = -5x + 7x`
`=> 6 = 2x`
`=> x= 3` (nhận)
Vậy `x = 1` hoặc `x =3`
Bài 3:
Cho `x= -2`
`=> y = f(-2) =4.(-2)^2 -9`
`=> y= f(-2) = 4.4 -9`
`=> y = f(-2) = 16 -9 = 7`
Cho `x= -1/2`
`=> y= f(-1/2)= 4.(-1/2)^2 -9`
`=> y= f(-1/2) = 4. 1/4 -9`
`=> y= f(-1/2) = 1 -9 =-8`
Vậy `f(-2) = 7; f(-1/2) = -8`
b) Để `f(x) =-1`
`=> 4x^2 -9 =-1`
`=> 4x^2 = -1 +9`
`=> 4x^2 = 8`
`=> x^2 = 2`
`=>` \(\left[ \begin{array}{l}x=\sqrt[]{2}\\x=-\sqrt[]{2}\end{array} \right.\)
Vậy `x= sqrt 2` hoặc `x= - sqrt 2`
Bài 4:
a) `f(x) = x( 1-2x) + (2x^2 -x + 4)`
`= x - 2x^2 + 2x^2 -x +4`
`= (x - x) - (2x^2 - 2x^2) +4`
`=4`
b) `g(x)= x(x-5) - x( x+2) + 7x`
`= x^2 - 5x - x^2 - 2x + 7x`
`=(x^2 - x^2) - (5x + 2x -7x)`
`= 0 -0`
`=0`
c) `h(x) = x(x-1) +1`
`= x^2 - x + 1`
