9.
Suatu segitiga siku-siku ABC dengan siku=siku di B. Dari titik B
ditarik garis lurus
memotong sisi AC di titik D. Sehingga BD tegak lurus AC. Panjang AD = (2x
+3) cm,
CD = (x+ 1) cm, dan BD = 2x cm. Ditentukan luas segitiga ABC = 39 cm2.
Tentukan keliling
segitiga ABC tersebut !
10. Diketahui : (3x – 4 y)2 + 22 xy = ax2 + by2
– cxy Hitunglah nilai ac - b
11. Jika f adalah
fungsi sehingga f(xy)= f(x–y) dan f(6) =1, maka f(–2) – f(5) =….
12. Tentukan nilai x dari :
2x3 12x2
27x
-27
81
. 27 . 9 =
3
13 Hitunglah nilai ( y + ⅔)2 yang
memenuhi persamaan
5 √(3x +2 +2√6x) = 125 y√3x + y√2
14. Selesaikan
(0,0081) 0,25 + {( 0,0016) -
(0,0625) } -4 = ….
15. Sederhanakan
y-1 + xy-2
1- x2 y-2
16. Jika an-4
= ¼ selesaikan
persamaan berikut :
(2a n)3 : (a n +1
)3
½ a 3n+3 a. a 4n
17. Tentrukan ni.lai y dari
15(24+1)(28+1)(216
+ 1)( 232+1)(264+1)(2128 + 1) = 2y -1
18. Tentukan
nilai a
8 (32 +1)(34+1)(38+1)(316 + 1)(
332+1)(364+1) = 1
6128a128 - 1
19. Jika x4 – x2 – x- 2 = 0 hitunglah x5 + x4 – x3 – 2x2 - 3x + 1
Solusi
soal aljabar No.9:
39 = ½ [ 2x {(x +1)+(2x + 3)} ]
39 = ½ [2x (3x +4)]
39 = 3x2 + 4x
3x2 + 4x – 39 = 0
3x2 – 9 x + 13x – 39 = 0
(3x2 – 9 x) + (13x – 39) = 0
3x (x – 3) + 13 (x – 3) = 0
(3x+ 13) (x – 3) = 0
X yang
memenuhi adalah x = 3
Sehingga AD
= 9 cm, CD = 4 cm, BD = 6 cm Serta AC = 13 cm
AB = √( 62
+ 92 )
AB = √( 36 +
81 )
AB = √117
BC = √( 62
+ 42 )
BC = √( 36 +
16 )
BC = √52
Jadi,
Keliling segitiga ABC = (√117 + √52 + 13 ) cm
Solusi
soal aljabar No.10:
(3x – 4 y)2 + 22 xy = ax2 + by2 – cxy
9x2 – 24 xy + 16y2 + 22 xy = ax2 + by2
– cxy
9x2 + 16y2 - 2 xy = ax2 + by2
– cxy
a = 9, b =16
, c = 2
Jadi, ac – b = 2
Solusi soal aljabar No.11:
f(xy)= f(x–y)
Diketahui f(6) =1
f(6) = f(3.2) =f(3–2)= 1, maka f(1)
= 1
f(2) = f(2.1) =f(2–1)=f(1)=1
f(3) = f(3.1) = f(3–1) = f(2)
= 1
f(4) = f(4.1) = f(4–1) = f(3)
= 1
f(5) = f(5.1) = f(5–1) = f(4) = 1
Selanjutnya
f(–2) = f(2(–1)) = f(2– (–1)) = f(3)
= 1
Jadi f(–2) – f(5) = 1 – 1 = 0
Solusi
soal aljabar No.12:
2x3 12x2
27x
-27
81
. 27 . 9 =
3
8x3 36x2 54x -27
3
. 3 . 3 =
3
8x3 + 36x2 + 54x + 27 = 0
( 2x + 3)3 = 0
2x + 3 = 0
X = - 3/2
Solusi
soal aljabar No.13:
5 √(3x +2 +2√6x) = 125 (y√3x
+ y√2)
5 √(3x +2 +2√6x) = 5 3y√3x + 3y√2
√(3x +2 +2√6x) = 3y√3x + 3y√2
√(3x +2 +2√6x) = 3y(√3x + √2)
√( √3x +√2)2 = 3y(√3x + √2)
√3x +√2 = 3y(√3x + √2)
3y = √3x + √2
√3x +√2
3y = 1
y = ⅓
( y + ⅔)2
= (⅓ + ⅔ )2 = 1
Solusi
soal aljabar No.14:
(0,0081) 0,25 + {( 0,0016) -(0,25)2
} -4 = ….
(0,34) ¼ + ( 0,
24) ¼ = 0,3 + 0,2
= 0,5
Solusi
soal aljabar No.15:
Petunjuk : Kalikan penyebut dan pembilang dengan huruf
berpangkat negative terbesar, kemudian pergunakan sifat
dan
y-1 + xy-2 =
y-1 + xy-2 .
y2
1- x2 y-2
1- x2 y-2
y2
= y
+ x
y2 – x2
= y
+ x
(y + x )(y – x)
= .
1 .
y + x
Solusi
soal aljabar No.16:
(2a n)3 : (a n +1
)3 = 8a 3n :
a 3n +3
½ a 3n+3 a. a 4n
½ a 3n+3 a. a 4n+
1
=
16 a-3 : a - n +1
=
16 a-3. a n - 1
=
16 an - 4
= 4
Solusi soal aljabar No.17:
Ingat
a2-b2 = (a-b)(a+b)
15( 24 +1) = 28
-1
(24 -1)( 24 +1) = 28
-1
(28 -1)( 28 +1) = 216
- 1
(216 – 1)( 216 + 1) =
232 - 1
(232 – 1)( 232 + 1) = 264
- 1
(264 – 1)( 264
+ 1 ) = 2128 - 1
(2128 – 1)( 2128
– 1) = 2256 – 1
Jadi,
y = 256
Solusi soal aljabar No.18:
8 (32 +1) = (32 -1) (32 +1) = 34 -1
(34 -1)( 34 +1) = 38 -1
(38 -1)( 38 +1) = 316 - 1
(316 – 1)( 316 + 1) = 332 - 1
(332 – 1)( 332 + 1) = 364 - 1
(364 – 1)( 364 + 1 ) = 3128 - 1
8 (32
+1)(34+1)(38+1)(316 + 1)( 332+1)(364+1)
= 6128a128 - 1
3128 – 1 = 6128a128 - 1
3128 – 1 = (6a)128 - 1
3128 = (6a)128 – 1+ 1
3128 = (6a)128
6a = 3
a = ½
Solusi
soal aljabar No.19:
x5 + x4 – x3 – 2x2 - 3x + 2 = x5 + x4 – x3 – x2 - x2 - 2x - x + 1
= x5– x3 – x2 - 2x + x4 - x2 - x + 1
= x5– x3 – x2 –2x + x4 - x2 - x+ 1
= x5– x3 – x2 – 2x + x4 - x2 - x+ 1 + 2 - 2
= x5– x3 – x2 – 2x + x4 - x2 – x- 2 + 3
=
x (x4 – x2 – x-2) + (x4 – x2 – x- 2 ) + 3
=
x (0) + (0) + 3
=
3
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