Nov 8, 2022
Loading...
seen from Russia
seen from United States
seen from Singapore
seen from Malaysia
seen from India
seen from China
seen from Ireland
seen from Australia
seen from United States
seen from United Kingdom
seen from United States
seen from China
seen from Australia
seen from China
seen from Russia
seen from China
seen from Malaysia
seen from Denmark
seen from United States
seen from Australia
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
Free to watch • No registration required • HD streaming
Soal dan Penyelesaian Persamaan Nilai Mutlak Kelas X SMA Semester 1
Ini soal matematika bab persamaan nilai mutlak.
Tentukan HP dari persamaan:
|x +5| = 3
|2x-3| = 5
|x+1| + 2x = 7
|3x+4| = x-8
|5- 2/3x| – 9 = 8
|x-7| + |2x-4|=5
|2x+4| – |3-x|= -1
|5x+10| = -|x2+2x|
Penyelesaian:
Penyelesaiannya ada 2 cara.. Terserah mau memakai cara yang mana hehe
Caranya Tika contohkan untuk nomor 1 saja ya! Selanjutnya Tika akan menggunakan cara yang Tika suka. Hehe
Cara pertama:
|x+5| = 3
Kedua ruas dikuadratkan karena mutlaknya bisa bernilai positif atau negatif, menjadi
<=> (x+5)2 = 32
<=> x2+10x+25 = 9
<=> x2+10x+25-9 = 0
<=> x2+10x+16 = 0
<=> (x+8) = 0 V (x+2) = 0
<=> x = -8 V x = -2
Dicek:
Untuk x = -8, ruas kiri = |x+5| = |-8+5| = |-3| = 3 = Ruas kanan
Jadi, x = -8 memenuhi
Untuk x = -2, ruas kiri = |x+5| = |-2+5| = |3| = 3 = ruas kanan
Jadi, x = -2 memenuhi
Jadi, HP = {-2, -8}
Cara kedua:
Untuk mutlak bernilai positif:
(x+5) = 3
<=> x = 3-5
<=> x = -2
Untuk mutlak bernilai negatif:
-(x+5) = 3
<=> -x-5 = 3
<=> -x = 3 + 5
<=> -x = 8
<=> x = -8
Dicek:
Untuk x = -8, ruas kiri = |x+5| = |-8+5| = |-3| = 3 = Ruas kanan
Jadi, x = -8 memenuhi
Untuk x = -2, ruas kiri = |x+5| = |-2+5| = |3| = 3 = ruas kanan
Jadi, x = -2 memenuhi
Jadi, HP = {-2, -8}
Hehe… sama, kan??
|2x-3| = 5
Untuk mutlak bernilai positif:
(2x-3) = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
Untuk mutlak bernilai negatif:
-(2x-3) = 5
<=> -2x+3 = 5
<=> -2x = 5-3
<=> -2x = 2
<=> x = -1
Dicek:
Untuk x = 4, ruas kiri = |2x-3| = |2.4 – 3| = |8 – 3| = |5| = 5 = ruas kanan
Jadi, x = 4 memenuhi
Untuk x = -1, ruas kiri = |2x-3| = |2.(-1) – 3| = |(-2)-3| = |-5| = 5 = ruas kanan
Jadi, x = -1 memenuhi
Jadi, HP = {-1, 4}
|x+1| + 2x = 7
Untuk mutlak bernilai positif:
(x+1) + 2x = 7
<=> 3x = 6
<=> x = 2
Untuk mutlak bernilai negatif:
-(x+1) + 2x = 7
<=> -x-1+2x = 7
<=> x = 8
Dicek:
Untuk x = 2, ruas kiri = |x+1| + 2x = |2+1| + 2.2 = |3| + 4 = 3 + 4 = 7 = ruas kanan
Jadi, x = 2 memenuhi
Untuk x = 8, ruas kiri = |x+1| + 2x = |8+1| + 2.8 = |9| + 16 = 9 + 16 = 25 tdk sama dengan ruas kanan
Jadi, x = 8 tidak memenuhi
Jadi, HP = {2}
|3x+4| = x-8
Untuk mutlak bernilai positif:
(3x+4) = x-8
<=> 2x = -12
<=> x = -6
Untuk mutlak nilai negatif:
-(3x+4) = x-8
<=> -3x-4 = x-8
<=> -4x = -4
<=> x = 1
Dicek:
|3x+4| = x-8
ó |3x+4| – x = -8
Untuk x = -6, ruas kiri = |3x+4| – x = |3.(-6) + 4| – (-6) = |-14| + 6 = 14 + 6 = 20 tdk sama dengan ruas kanan
Jadi, x = -6 tidak memenuhi
Untuk x = 1, ruas kiri = |3x+4| – x = |3.1 + 4| – 1 = |7| – 1 = 7 – 1 = 6 tdk sama dengan ruas kanan
Jadi, x = 1 tidak memenuhi
Jadi, HP = {}
|5- 2/3x| – 9 = 8
Untuk mutlak bernilai positif:
(5- 2/3x) – 9 = 8 |x3
<=> (15 – 2x) – 27 = 24
<=> -2x = 36
<=> x = -18
Untuk mutlak bernilai negatif:
-(5- 2/3x) – 9 = 8 |x3
<=> -(15 – 2x) – 27 = 24
<=> -15+2x-27 = 24
<=> 2x = 66
<=> x = 33
Dicek:
Untuk x = -18, ruas kiri = |5 -2/3x| – 9 = |5.(-18)| – 9 = |5+12| – 9 = |17| – 9 = 17 – 9 = 8 = ruas kanan
Jadi, x = -18 memenuhi
Untuk x = 33, ruas kiri = |5 -2/3x| – 9 = |5.(33)| – 9 = | 5-22| – 9 = |-17| – 9 = 17 – 9 = 8 = ruas kanan
Jadi, x = 33 memenuhi
Jadi, HP = {-18, 33}
|x-7| + |2x-4|= 5
Untuk mutlak nilai positif:
(x-7) + (2x-4) = 5
<=> 3x = 16
<=> x = 16/3
Untuk mutlak nilai negatif:
-(x-7) + (-(2x-4)) = 5
<=> -x+7-2x+4 = 5
<=> -3x = -6
<=> x = 2
Dicek:
Untuk x = 16/3, ruas kiri = |x-7| + |2x-4| = |16/3 – 7| + |2.16/3 – 4| = |-5/3| + |20/3| = 5/3 + 20/3 = 25/3 tdk sama dengan ruas kanan
Jadi, x = 16/3 tidak memenuhi
Untuk x = 2, ruas kiri = |x-7| + |2x-4| = |2-7| + |2.2-4| = |-5| + |0| = 5 + 0 = 5 = ruas kanan
Jadi, x = 2 memenuhi
Jadi, HP = {2}
|2x+4| – |3-x|= -1
Untuk mutlak bernilai positif:
(2x+4) – (3-x)= -1
<=> 2x+4-3+x = -1
<=> 3x = -2
<=> x = -2/3
Untuk mutlak bernilai negatif:
-(2x+4) – (-(3-x)) = -1
<=> -2x-4+3-x = -1
<=> -3x = 0
<=> x = 0
Dicek:
Untuk x = -2/3, ruas kiri = |2x+4| – |3-x| = |2.(-2/3)+4| – |3-(-2/3)| = |8/3| – |11/3| = 8/3 – 11/3 = -3/3 = -1 = ruas kanan
Jadi, x = -2/3 memenuhi
Untuk x = 0, ruas kiri = |2x+4| – |3-x| = |2.0 + 4| – |3 – 0| = |4| – |3| = 4 – 3 = 1 tdk sama dengan ruas kanan
Jadi, x = 0 tidak memenuhi
Jadi, HP = {-2/3}
|5x+10| = -|x2+2x|
<=> |5x+10| + |x2+2x| = 0
Untuk mutlak bernilai positif:
(5x+10) + (x2+2x) = 0
<=> x2 + 7x + 10 = 0
<=> (x+5) = 0 V (x+2) = 0
<=> x = -5 V x = -2
Untuk mutlak bernilai negatif:
-(5x+10) + (-(x2+2x)) = 0
<=> -5x-10- x2-2x = 0
<=> – x2 -7x-10 = 0 |x(-1)
<=> x2 +7x+10 = 0
<=> (x+5) = 0 V (x+2) = 0
<=> x = -5 V x = -2
Dicek:
Untuk x = -5, ruas kiri = |5x+10| + |x2+2x|= |5.(-5)+10| + |(-5)2 + 2.(-5)| = |-15| + |15| = 15 + 15 = 30 tdk sama dengan ruas kanan
Jadi, x = -5 tidak memenuhi
Untuk x = -2, ruas kiri = |5x+10| + |x2+2x|= |5.(-2)+10| + |(-2)2 + 2.(-2)| = |0| + |0| = 0+0 = 0 = ruas kanan
Jadi, x = -2 memenuhi
Jadi, HP = {-2}
Semoga membantu Jika ada yang tidak paham tanya saja yaaa
English Debate Competition Team from @smait_ruhuljadid ~ . . Alhamdulillah.. All praises be to Allah, because of Allah's help, they can speech well. . . They battled Penabur High School students. The score between Ruhul Jadid and Penabur is 51:38. I think, it is not bad.., because it is the first time they join English Debate Competition. Moreover, we only practiced last night.✊ . . Our spirit is in Qs. Al Imran: 139. ("Dont feel weak and dont be sad because you are in the highest level if you have faith to Allah"). . . Next year, Ruhul Jadid team must be better than today, fighting! ALLAHU AKBAR! ✊✊✊ #englishdebate #competition #SMAITRuhulJadid #Kelas10