No 9 sama 10 gimana caranya

[tex]9. |2x + 3| > |x - 4| \\ \sqrt{(2x + 3) ^{2} } > \sqrt{(x - 4)^{2} } \\ (2x + 3)^{2} - (x - 4)^{2} > 0 \\ (2x + 3 + x + 4) \: (2x + 3 - x - 4) > 0 \\ (3x + 7) \: (x - 1) > 0 \\ x = - \frac{7}{3} \: atau \: x = 1 \\ \\ 10. |2x| \leqslant 2 \\ \sqrt{(2x)^{2} } \leqslant 2 \\(2x) ^{2} \leqslant {2}^{2} \\ (2x)^{2} - 2^{2} \leqslant 0 \\ (2x + 2) \: (2x - 2) \leqslant 0 \\ x = - 1 \: atau \: x = 1 \\ [/tex]

Jawabannya:

9. * |2x+3|>|x-4|
= 2x+x > -3-4
= 3x> -7
= x> -7/3
* |2x-3|>|x+4|
= 2x-x>3+(-4)
= x>-1
= x<1

Jadi HPnya:{ -7/3 < x < 1 } --- E

10. |2x|≤2
2x≤2
x≤2/2
x≤1. ---- C

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