t3 bab 6
DESCRIPTION
MATH T3 - BAB 6 UNGKAPAN ALGEBRATRANSCRIPT
![Page 1: T3 BAB 6](https://reader036.vdocuments.site/reader036/viewer/2022082422/563db803550346aa9a8fbf8f/html5/thumbnails/1.jpg)
UNGKAPAN
ALGEBRA
BAB 6
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RECALL
12 x a =
27a x a =
33a x 2a =
42 x (–a) =
5–2 x 4a =
6–2a x (–3a) =
7p x (–q) =
82p x 4q =
9–p x 5q =
10 p x 3pq
=11 q x (– 3pq)
=12 –5pq x (–
2p)=
2a
7a2
6a2
–2a
–8a
6a2
–pq
8pq
–5pq
3p2q
–3pq2
10p2q
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+
a(b+c) =
ab +ac x
x
3(k+5) =
3k + 15
xx
5(y – 5) =
5y – 25
xx
–
–3(k+5) =
– 3k –15
–5(y – 5) =
– 5y
+ 25
xx
xx1
3
2
4
KEMBANGAN l
Kembangkan :
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+
(a+b)(c+d) = ac + ad
x x
(p+q)(p–5) =
p2–5p x
x
–
1
+ bc
+ bd x x
x
+pq
–5q x
(h–4)(k–2) =
hk –2h x
x2 x
– 4k
+8x
KEMBANGAN lI
Kembangkan :
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+
(2x+3)(x –4) =
2x2
–8x x
x
–
3 x
+3x
–12 x
= 2x2–5x
–12
(3h–6)(2h –5) =
6h2
–15h
x
x4 x
–12h
+30x
= 6h2–27h
+30
KEMBANGAN lI
Kembangkan :
![Page 6: T3 BAB 6](https://reader036.vdocuments.site/reader036/viewer/2022082422/563db803550346aa9a8fbf8f/html5/thumbnails/6.jpg)
Masuk dalam kurungan
2x – 8 =
– 4
1
2
2
PEMFAKTORAN I
Faktorkan :
2x – 8
x
( x – 4)Duduk luar
!!
6x2 + 9x =
+ 3x
2
3
3x
2x2
(2x + 3)
6x2 + 9x x2x + 3
= a(b+c)
ab +
ac
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12xy – 8y =
– 2y
3
4
4y
PEMFAKTORAN I
Faktorkan :
3xy
(3x – 2)
6x2 + 9x =
+ 3x
4
3
3x
2x2
(2x + 3)
6x2 + 9x x2x + 3
12xy – 8yy
– 2 3x
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x2 – 16
1
= x2
PEMFAKTORAN Il
Faktorkan :
– 42
2
a2 – b2 = (a+b)
(a–b)
= (x+4)
(x–4 )
25 – m2
= 52
– m2
= (5+m)
(5–m )
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9x2 – 16
3
=32
x2
PEMFAKTORAN Il
Faktorkan :
– 42
4
a2 – b2 = (a+b)
(a–b)
= (3x+4)
(3x–4 )
25 – 81m2
= 52
– 92m2
= (5+9m)
(5–9m )
=(3x)2
– 42 = 52
– (9m)2
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x2 + 5x + 6
1
PEMFAKTORAN IlI
Faktorkan :
x
= (x+2)
(x+3 )
+x
23
x2
6 +5x
2x3x++
+
+
( ) ( )
D T B D B T
1 x 62 x 3
x +
ax2 + bx + c
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x2 +4x –12
2
PEMFAKTORAN IlI
Faktorkan :
x
= (x+6)
(x–2 )
+x
62
x2
12 +4x
6x2x––
+
–
( ) ( )
D T B D B T
1 x 122 x 63 x 4
x +
ax2 + bx + c
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2x2 +3x –20
3
PEMFAKTORAN IlI
Faktorkan :
2x
= (2x–5)
(x+4 )
–x
54
2x2
20 +3x
5x8x++
–
–
( ) ( )
D T B D B T
1 x 202 x 104 x 5
x +
ax2 + bx + c
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ax +bx +ay +by
1
PEMFAKTORAN IV
Faktorkan :
= x
y(a+b)
+ (a+b)=
(a+b)(x+y)
3h +hk –6–2k
2
= h
2(3+k)
– (3+k)=
(3+k)(h–2)
ax +bx +ay +by = (a+b)(x+y)
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1
PENAMBAHAN DAN PENOLAKAN PECAHAN
ALGEBRA
Permudahkan :
2
8y5
4x
8x2
8y5
8y5x2
v35
u43
uv12v9
uv12u20
uv12u20v9
x 2
x 2x 4u
x 3v
x 3v x 4u
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1kk31k
( )
3
PENAMBAHAN DAN PENOLAKAN PECAHAN
ALGEBRA
Permudahkan :
4
k31
1k2
1kk3k6
2m6n
mn23
nm6m9
2nm6
n2
2
nm6nm9
2
2 1kk3
1kk6
1kk31k5
x 3k
x 3k
x (k+1)
x (k+1)
x 3m
x 3m
x n
x n
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5
PENAMBAHAN DAN PENOLAKAN PECAHAN
ALGEBRA
Permudahkan :
2m62m
m21
2m6m3
2m62m
2m62mm3
2m62m2
2m31m
x 3m
x 3m
( )
1 1
2
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1
PENDARABAN DAN PEMBAHAGIAN PECAHAN
ALGEBRA
Permudahkan :
2
c7b3
5a4
ab12
c35
x3zx
2yyz4
zxyz4 2yx3
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3
PENDARABAN DAN PEMBAHAGIAN PECAHAN
ALGEBRA
Permudahkan :
4
vw3
wuv4
2uv4 2w3
wuv4
w3v
bab3
3a2a5
baa5
3a2b3
3a2a5
b3ba
![Page 19: T3 BAB 6](https://reader036.vdocuments.site/reader036/viewer/2022082422/563db803550346aa9a8fbf8f/html5/thumbnails/19.jpg)
5
PENDARABAN DAN PEMBAHAGIAN PECAHAN
ALGEBRA
Permudahkan :
22
2
2
2
kh1h
khkhh
khk
1hh
2
2
khkhh
1h
kh2
22
1h1h
khkh
khh 1hk