pertidaksamaan non linear

1. Pertidaksamaan Non Linear
A. Pertidaksamaan dengan pangkat lebih dari
satu
Tentukan himpunan penyelesaian dari
pertidaksamaan berikut
(1) x2
– 4x ≤ –3
(2) x3
– 3x2
– 10x > 0
(3) –x2
+ 6x – 9 < 0
Selesaian:
(1) x2
– 4x ≤ –3
 x2
– 4x + 3 ≤ 0
 x2
– 4x + 3 = 0
 (x – 3) (x – 1) = 0
 x – 3 = 0 x – 1 = 0
 x = 3 x = 1
Karena lambang “≤” maka bulatan arsir dan
penyelesaian negatif
garis bilangan:
III II I
1 3
uji nilai fungsi: (misal)
interval I : x ≥ 3 maka ambil x = 4
interval II : 1 ≤ x ≤ 3 maka ambil x = 2
interval III : x ≤ 1 maka ambil x = 0
interval x x2
– 4x + 3 ket
x ≥ 3 4 42
– 4(4) + 3 = 3 +
1 ≤ x ≤ 3 2 22
– 4(2) + 3 = –1 –
x ≤ 1 0 02
– 4(0) + 3 = 3 +
jadi garis bilangan
+ – +
1 3
1 ≤ x ≤ 3
Jadi Hp = {x | 1 ≤ x ≤ 3, x  R}
(2) x3
– 3x2
– 10x > 0
 x3
– 3x2
– 10x = 0
 .... (......................) = 0
 .... (...........)(...........) = 0
 x = 0 ......... = 0 ......... = 0
 x = ....... x = ....... x = .......
Karena lambang “>” maka bulatan ............. dan
penyelesaian .................
interval x x3
– 3x2
– 10x ket
jadi garis bilangan
..... ..... .....
............................................................................
Jadi Hp = {x | ............................................, x  R}
(3) –x2
+ 6x – 9 < 0
 ....................... =0
 (.............) (.............) = 0
 ............. = 0 ............. = 0
 x = ........ x = ........
Karena lambang “<” maka bulatan ............. dan
penyelesaian .................
Kelas:
Nama:

2. interval x –x2
+ 6x – 9 ket
jadi garis bilangan
.....
............................................................................
Jadi Hp = {x | ............................................, x  R}
B. Pertidaksamaan yang memuat bentuk
pecahan
Tentukan himpunan penyelesaian dari
pertidaksamaan berikut
(1)
𝑥−6
𝑥+2
≤ 3
(2)
3𝑥2+5𝑥−8
𝑥2+𝑥−6
≥ 3
Selesaian:
INGAT MINTA BANTUAN PERSAMAAN
(1)
𝑥−6
𝑥+2
≤ 3
⟺
𝑥−6
𝑥+2
− ...... ≤ 0
⟺
𝑥−6
𝑥+2
−
3(............)
(............)
≤ 0
⟺
.............................
............
≤ 0
⟺
...............
............
≤ 0
bantuan:
...............
............
= 0
pembilang: penyebut:
................. = 0 ................. = 0
 ......... = .......  ......... = .......
 ......... = .......
 ......... = .......
Ingat
0
1
= ......
1
0
= ......
Apakak penyebut boleh nol? .............
Karena lambangnya adalah “≤” maka bulatan
.............., penyelesaian .................,
tapi penyebut ≠ 0
interval x ket
jadi garis bilangan:
............................................................................
............................................................................
............................................................................
Jadi Hp = {x | ............................................, x  R}
(2)
3𝑥2+5𝑥−8
𝑥2+𝑥−6
≥ 3
⟺
3𝑥2+5𝑥−8
𝑥2+𝑥−6
− ....... ≥ 0
⟺
3𝑥2+5𝑥−8
𝑥2+𝑥−6
−
3(……….............)
(.......……..........)
≥ 0
⟺
………………….............................
..………........
≥ 0
⟺
...............
............
≥ 0
bantuan:
...............
............
= 0
pembilang: penyebut:
................. = 0 ...................... = 0
 ......... = .......  (............)(...........) = 0
 ......... = .......  ...... = ...... ....... = .......
 ......... = .......  ...... = ...... ....... = .......
pindah ruas
samakan penyebut

3. Karena lambangnya adalah “≥” maka bulatan
.............., penyelesaian .................,
tapi ............................
interval x ket
jadi garis bilangan:
............................................................................
............................................................................
............................................................................
Jadi Hp = {x | ............................................, x  R}
Kerjakan soal berikut:
Tentukan himpunan penyelesaian berikut
a)
4𝑥−3
𝑥+2
> 0
b)
3𝑥+6
6𝑥−3
≥ 1
c)
2𝑥2−7𝑥+6
𝑥2−7𝑥+10
< 0
d)
𝑥2+11𝑥+14
3𝑥2+5𝑥+2
≤ −1
C. Pertidaksamaan yang memuat bentuk
akar
bentuk umum:
√𝑓(𝑥) > √𝑔(𝑥) ⟹ 𝑓(𝑥) > 𝑔(𝑥)
dengan syarat f(x) ≥ 0
g(x) ≥ 0
Tentukan himpunan penyelesaian dari
pertidaksamaan berikut
(1) √3𝑥 − 4 < √ 𝑥 + 2
(2) √4𝑥 − 8 > 4
(3) √
2𝑥−6
𝑥+3
< 3
(4) √𝑥2 − 4𝑥 − 21 ≥ 𝑥 − 9
Selesaian:
(1) √3𝑥 − 4 < √ 𝑥 + 2
syarat:
 3x – 4 ≥ 0
 ............................
 ............................
 x + 2 ≥ 0
 ............................
 √3𝑥 − 4 < √ 𝑥 + 2
 ........................................
 ........................................
 ........................................
 ........................................
 ........................................
garis bilangan:
S1
............................................................................
S2
............................................................................
S3
............................................................................
HP
............................................................................
............................................................................
Jadi Hp = {x | ............................................, x  R}
kuadratkan
kedua ruas

4. (2) √4𝑥 − 8 > 4
syarat:
 4x – 8 ≥ 0
 ............................
 ............................
 ............................
 √4𝑥 − 8 > 4
 ........................................
 ........................................
 ........................................
 ........................................
 ........................................
garis bilangan:
S1
............................................................................
S2
............................................................................
HP
............................................................................
............................................................................
Jadi Hp = {x | ............................................, x  R}
(3) √
2𝑥−6
𝑥+3
< 3 ⟺
√2𝑥−6
√𝑥+3
< 3
syarat:

2𝑥−6
𝑥+3
≥ 0
bantuan:
...............
............
= 0
pembilang: penyebut:
................. = 0 ................. = 0
 ......... = ....... ......... = .......
 ......... = ....... ingat penyebut ≠ 0
 ......... = .......
 √
2𝑥−6
𝑥+3
< 3
 ........................................
 ........................................
 ........................................
 ........................................
bantuan:
...............
............
= 0
pembilang: penyebut:
................. = 0 ................. = 0
 ......... = ....... ......... = .......
 ......... = ....... ingat penyebut ≠ 0
garis bilangan:
S1
............................................................................
S2
............................................................................
HP
............................................................................
............................................................................
Jadi Hp = {x | ............................................, x  R}
Jadi Hp = {x | ............................................, x  R}
kuadratkan
kedua ruas

5. (4) √𝑥2 − 4𝑥 − 21 ≥ 𝑥 − 9
syarat:
 𝑥2
− 4𝑥 − 21 ≥ 0
 ............................
 ............................
 ............................
 √𝑥2 − 4𝑥 − 21 ≥ 𝑥 − 9
 ........................................
 ........................................
 ........................................
 ........................................
 ........................................
 ........................................
 ........................................
garis bilangan:
S1
............................................................................
S2
............................................................................
HP
............................................................................
............................................................................
Jadi Hp = {x | ............................................, x  R}
Tentukan himpunan penyelesaian dari
pertidaksamaan berikut
a) √6𝑥 − 2 ≤ √ 𝑥 + 8
b) 2 ≥ √ 𝑥 + 10
c) √
5
4
𝑥 +
15
12
< √
3
2
𝑥 +
6
4
d) √𝑥2 − 8𝑥 + 16 > 2
D. Pertidaksamaan yang memuat nilai mutlak
Ingat kembali sifat-sifat pertidaksamaan nilai
mutlak
|x| < p  ................................
|x| ≤ p  ................................
|x| > p  ................................
|x| ≥ p  ................................
|x| < |y|  ................................
|x| ≤ |y|  ................................
|x| > |y|  ................................
|x| ≥ |y|  ................................
Tentukan himpunan penyelesaian dari
pertidaksamaan berikut
(1) |2x – 3| ≥ 7
(2) |3x – 2| – |–5| > 0
(3) |4x + 1| < 7
(4) |5x – 1| ≥ |5x + 3|
Selesaian:
(1) |2x – 3| ≥ 7
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
Jadi Hp = {x | ............................................, x  R}
(2) |2x – 3| ≥ 7
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
Jadi Hp = {x | ............................................, x  R}
(3) |4x + 1| < 7
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
Jadi Hp = {x | ............................................, x  R}

6. (4) |5x – 1| ≥ |5x + 3|
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
 .....................................................
Jadi Hp = {x | ............................................, x  R}
Tentukan himpunan penyelesaian dari
pertidaksamaan berikut
a) |3x + 1| ≤ 6
b) |
𝑥
4
− 1| ≤ 6
c) |4x – 6| – 4 |x + 4| > 0