4. (x – 2 = 1 X-2 =1 x = 1+2. X-2=-1 x= -1+2 x = ….. x = ….
1. 4. (x – 2 = 1 X-2 =1 x = 1+2. X-2=-1 x= -1+2 x = ….. x = ….
Jawaban:
(x-2=1
X-2=1
x=1+2
X-2=-1
X=-1+2
x=1
Jadi jawabannya adalah 1
semoga membantu
2. 2∣x−1∣≤∣x+2∣[tex]2∣x−1∣≤∣x+2∣[/tex]2∣x−1∣≤∣x+2∣
Jawaban:
428#@72:$@9264388#@kamu itu sanga tidak pintar kon
3. Jika x^1/2 + x^-1/2 = 2 Maka x^2 +x^-2= …?
Diketahui [tex]x^{\frac{1}{2}}+x^{-\frac{1}{2}}=2[/tex]
Kuadratkan kedua ruas,
=> [tex](x^{\frac{1}{2}}+x^{-\frac{1}{2}}-2)^{2}=0[/tex]
=> [tex](x^{\frac{1}{2}}+x^{-\frac{1}{2}}-2)(x^{\frac{1}{2}}+x^{-\frac{1}{2}}-2)=0[/tex]
=> [tex]x+1-2x^{\frac{1}{2}}+1+x^{-1}-2x^{-\frac{1}{2}}-2x^{\frac{1}{2}}-2x^{-\frac{1}{2}}+4=0[/tex]
=> [tex]x-4x^{\frac{1}{2}}+x^{-1}-4x^{-\frac{1}{2}}+6=0[/tex]
=> [tex]x+x^{-1}-4(x^{\frac{1}{2}}+x^{-\frac{1}{2}})=-6[/tex]
=> [tex]x+x^{-1}-8=-6[/tex]
=> [tex]x+x^{-1}=2[/tex]
Sekarang kuadratkan lagi,
=> [tex](x+x^{-1}-2)^{2}=0[/tex]
=>
[tex](x+x^{-1}-2)(x+x^{-1}-2)=0[/tex]
=> [tex]x^{2}+1-2x+1+x^{-2}-2x^{-1}-2x-2x^{-1}+4=0[/tex]
=> [tex]x^{2}+x^{-2}-4x-4x^{-1}+6=0[/tex]
=> [tex]x^{2}+x^{-2}-4(x+x^{-1})=-6[/tex]
=> [tex]x^{2}+x^{-2}-8=-6[/tex]
=> [tex]x^{2}+x^{-2}=2[/tex]
4. jika x^1/2+x^-1/2=2 maka nilai dari x^2+x^-2
x^1/2 + x^-1/2 = 2
(√x + 1/√x)² = 2²
x + 2 + 1/x = 4
x + 1/x = 4 – 2
x + 1/x = 2
(x + 1/x)² = 2²
x² + 1/x² + 2 = 4
x² + 1/x² = 4 – 2
x² + 1/x² = 2
⏺x^2 + x^-2 = 2
5. 1. jika f(x) = x+1 dan g(x)= x^2 maka (fog) (x) a. (x+1)^2b (x-1)^2c x^2 + 1d x^2 -1 e x^22. jika f(x) = 2^x dan g(x) =x^2 maka (gof) (x) a (2^x)^2b 2^x^2c 4xd 2x^2e 2^x + x^2
semoga membantuuuuuuu1.C
2.D
jawabannya klo gk slh itu
6. Hasil bagi dari pembagian suku banyak (x^6-1):(x²-1) adalah… a. (x^2+x+1) (x^2+x-1) b. (x^2+x+1) (x^2-x+1) c. (x^2+x+1) (x^2-x-1) d. (x^2-x+1) (x^2+x-1) e. (x^2+x-1) (x^2-x-1) beserta caranya ya..
Penjelasan dengan langkah-langkah:
dilampirkan pada gambar…
7. Jika x^1/2 + x^-1/2 = 2 Nilai x^2 + x^-2 adalah
x^1/2 + x^-1/2 = 2
(x^1/2 + x^-1/2)^2 = 2^2
(x^1/2)^2 + 2 . x^1/2 . x^-1/2 + (x^-1/2)^2 = 4
x + 2 + x^-1 = 4
x + x^-1 = 2
(x + x^-1)^2 = 2^2
x^2 + 2 . x . x^-1 + x^-2 = 4
x^2 + 2 + x^-2 = 4
x^2 + x^-2 = 2
Ingat : (a + b)^2 = a^2 + 2ab + b^2
8. 1/2 x 1/2 x 1/2 x 1/2 x 1/2 =
1/2 pangkat 5=0.01325
kalo gk salah1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32
9. jika X^1/2 + x^-1/2 = 2 , maka nilai dari X^2+x^-2 adalah
Penjelasan dengan langkah-langkah:
^ artinya pangkat
x^(1/2) + x^(-1/2) = 2
(x^(1/2) + x^(-1/2))² = 2²
(x^(1/2))² + 2 . x^(1/2) . x^(-1/2) + (x^(-1/2))² = 4
x + 2 + x^(-1) = 4
x + x^(-1) = 4 – 2
x + x^(-1) = 2
(x + x^(-1))² = 2²
x² + 2 . x . x^(-1) + x^(-2) = 4
x² + 2 + x^(-2) = 4
x² + x^(-2) = 4 – 2
x² + x^(-2) = 2
Detil jawaban
Kelas 9
Mapel 2 – Matematika
Bab 1 – Bilangan Berpangkat
Kode Kategorisasi : 9.2.1
10. 1] Jika f(x) = 1/2(a^x + a^-x) dan g(x) = 1/2(a^x – a^-x), maka f(x) . f(y) + g(x) . g(y) sama dengan ? 2] Jika x^1/2 + x^-1/2 = 2, maka nilai dari x^2 + x^-2 adalah ? 3] Jika x^1/2 + x^-1/2 = 2, maka x^3/2 + x^-3/2 sama dengan ?
E|<sp0|\|en
1]
f(x) = 1/2 (a^x + a^-x)
g(x) = 1/2 (a^x – a^-x)
•
f(x) . f(y) + g(x) . g(y)
= (1/2)^(a^x + a^-x) . (1/2)^ (a^y + a^-y) + (1/2)^(a^x + a^-x) . (1/2)^ (a^y + a^-y)
= 2 . (2^-1)^((a^x + a^-x) + (a^y + a^-y))
= 2^(1 – a^x – a^-x – a^y – a^-y)
2]
→ (a + b)² = a² + b² + 2ab
•
x^1/2 + x^-1/2 = 2
(x^1/2 + x^-1/2)² = 2²
x + 1/x + 2 = 4
x + 1/x = 2
•
(x + 1/x)² = 2²
x² + 1/x² + 2 = 4
x^2 + x^-2 = 2 ✔
3]
x^1/2 + x^-1/2 = 2
•
(x + 1/x)(x² + 1/x²) = x³ + x + 1/x + 1/x³
x³ + 1/x³ = (x + 1/x)(x² + 1/x²) – (x + 1/x)
•
x³ + 1/x³
= (x + 1/x)(x² + 1/x²) – (x + 1/x)
= 2 × 2 – 2
= 2 ✔
11. 1/2 x 1/2 x 1/2 x 1/2 ( 1 per 2 x 1 per 2 x ,1 per 2 x 1 per 2 ) nyatakan dalam bentuk bilangan berpangkat
Jawaban:
[tex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = { \frac{1}{2} }^{4}[/tex]
12. jika x^1/2 +x^-1/2=2 maka nilai x^2+x^-2 adalah…
x^1/2 +x^-1/2=2
aku sih nyoba2 aja
kalau x = 1 benar hasilnya 2
kalau x = yg lain diatas 1 hasilnya melebihi 2
Jadi di persamaan satunya x nya sama, x = 1 juga
hasilnya juga sama, 2
Maaf kalau salah
semoga membantu
Penyelesaian:
dapat diselesaikan dengan menggunakan segitiga pascal atau perpangkatan langsung, disini di kerjakan menggunakan perpangkatan langsung:
[tex] \sqrt{x} + \frac{1}{ \sqrt{x} } = 2 [/tex]
[tex]( \sqrt{x} + \frac{1}{ \sqrt{x} })^2 = 2^2 [/tex] (kedua ruas di kuadratkan)
[tex]x + \frac{1}{x} + 2 = 4[/tex]
[tex]x + \frac{1}{x} = 2[/tex]
[tex](x + \frac{1}{x})^2 = 2^2[/tex]
[tex]x^2 + \frac{1}{x^2} + 2 = 4[/tex] (kedua ruas di kuadratkan)
[tex]x^2 + \frac{1}{x^2} = 2[/tex]
13. jika x^1/2 + x^-1/2 = 2maka x^2 + x^-2 =…
semoga ini bermanfaat
(x^½) / 1 + 1 / (x^½) = 2
(x^½ * x^½)/(x^½) + 1/(x^½) = 2
(x^1 + 1) / x^1/2 = 2
x¹ + 1 = 2x^½
x¹ – 2x^½ + 1 = 0 , (x^½ – 1)² = 0
x^½ = 1 , (x^½)² = 1² , x = 1
jadi x^2 + x^-2 = 1^2 + 1^-2 = 1 + 1/1² = 2
jwbn : 2
14. jika x^1/2+ x^-1/2 = 2 maka nilai dari x^2+x^-2 adalah
x^1/2 + x^-1/2 = 2
(x^1/2 + x^-1/2)^2 = 2^2
x + 2 + x^-1 = 4
x + x^-1 = 2
(x + x^-1)^2 = 2^2
x^2 + 2 + x^-2 = 4
x^2 + x^-2 = 2
15. Jika tan 5⁰ = x, tentukan nilai sin 50⁰ …A.√2/2 ((x+1)/√(x^2-1))B.√2/2 ((x+1)/√(x^2+1))C.√2/2 ((x-1)/√(x^2+1))D.√2/3 ((x+1)/√(x^2+1))E.√2/3 ((x-1)/√(x^2+1))
Jawab:
B.√2/2 ((x+1)/√(x^2+1))
Penjelasan dengan langkah-langkah:
tan 5= x= x/1 » r = √(x²+1)
sin 5 = x/√(x²+1)
cos 5= 1/√(x²+1)
sin 50= sin (45+5)
= sin 45cos5 + cos 45sin 5
=√2/2 . 1/√(x²+1) + √2/2 . x/√(x²+1)
= √2 / 2√(x²+1) + x√2 / 2√(x²+1)
=( x√2 + √2) / 2√(x²+1)
= √2(x+1) / 2√(x²+1)
= √2/2 [ (x+1)/(x²+1) ]
16. 1. Lim x -> 1 x^2-1 / x^2+3x-2 2. Lim x -> -1 (x+1)(x-4)+2(x-1) / x^2+x-2
1) lim x>1 (x-1)(x+2)/ (x+1)(x+2)
=lim x>1 (x-1)/(x+2)
=(1-1)(1+2)
=0/3
=0
2) lim x-> -1 (x+1)(x-4)+2/(x+2)
=(-1+1)(-1-4)+2/(-1+2)
=(0)(-5)+2/(1)
=0+2/1
=2/1
=2
17. 1/2 x 1/2 x 1/2 x 1/2 x 1/2=
2×2×2×2×2=36
maaf kalo salah
18. berapa (x^2 +x^-2), jika (x^1/2+ x^-1/2)=2
(x^(1/2) + x^(-1/2)) = 2
=> (x^(1/2) + x^(-1/2))^2 = 2^2
=> x + 2 + x^(-1) = 4
=> x + x^(-1) = 2
(x + x^(-1))^2 = 2^2
=> x^2 + 2 + x^(-2) = 4
=> x^2 + x^(-2) = 2
19
. 1. ⅓(x-2)=⅔x-13/32. 2(½x+³/2)-7/2=³/2(x+1)-(½x+2)
cuma nomor satu yang dapat