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Pertemuan 2 dan 3 Alokasi Waktu : 6 x 50 menit

Pertemuan 2 dan 3 Alokasi Waktu : 6 x 50 menit. POKOK MATERI 1.2 HASIL KALI KHUSUS. HASIL KALI KHUSUS. a(c + d) = ac + ad (a – b)(a + b) = a 2 - b 2 (a + b)(a + b) = ( a + b) 2 = a 2 + 2ab + b 2 (a - b)(a - b) = ( a - b) 2 = a 2 - 2ab + b 2

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Pertemuan 2 dan 3 Alokasi Waktu : 6 x 50 menit

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  1. Pertemuan 2 dan 3AlokasiWaktu: 6 x 50 menit POKOK MATERI 1.2 HASIL KALI KHUSUS Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  2. HASIL KALI KHUSUS a(c + d) = ac + ad (a – b)(a + b) = a2 - b2 (a + b)(a + b) = ( a + b)2 = a2 + 2ab + b2 (a - b)(a - b) = ( a - b)2 = a2 - 2ab + b2 (x + a)(x + b) = x2 + (a+b)x + ab (ax + b)(cx + d) = acx2 + ( ad + bc)x + bd (a + b)(c + d) = ac + bc + ad + bd (a + b) (a + b) (a + b) = (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a - b) (a - b) (a - b) = (a - b)3 = a3 - 3a2b + 3ab2 - b3 ( a - b )( a2 + ab + b2) = a3 – b3 ( a + b )( a2 - ab + b2 = a3 + b3 (a + b + c)2 = a2 + b2 + c2 + 2ab + 2 ac + 2 bc Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  3. Bentuk – bentuk berikut dapat dibuktikan dengan perkalian (a – b)(a2 + ab + b2) = a3 – b3 (a – b )(a3 + a2 b + ab2 + b3 ) = a4 – b4 (a – b )(a4 + a3 b + a2b2 + ab3 + b4 ) = a5 – b5 (a – b )(a5 + a4 b + a3b2 + a2b3 + ab4 + b5 ) = a6 – b6 Dan seterusnya dan dapat dibuat secara umum: (a – b )(an-1 + an-2 b + an-3b2 + ..............+ abn - 2 + b n - 1 ) = a n – bn (14) n adalah sembarang bilangan positif ( 1, 2, 3, ..........) Dengan cara yang sama didapatkan: (a + b )(an-1 - an-2 b + an-3b2 - ..............- abn - 2 + b n - 1 ) = a n + bn ... (15) n adalah sembarang bilangan ganjil( 1, 3, 5, 7, ..........) Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  4. Untuk faktor (a+b) dan n ganjil (a + b)(a2 - ab + b2) = a3 + b3 (a + b )(a4 - a3 b + a2b2 - ab3 + b4 ) = a5 + b5 (a + b )(a6 – a5 b + a4b2 – a3b3 + a2 b4 – a b5 + b6 ) = a7+ b7 dan seterusnya dan dapat dibuat secara umum: (a + b )(an-1 - an-2 b + an-3b2 - ..............- abn - 2 + b n - 1 ) = a n + bn ............. (15) n adalah sembarang bilangan ganjil( 1, 3, 5, 7, ..........) Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  5. Contoh 1 (3x +5y)2 = (3x)2 + 2(3x)(5y) + (5y)2 = 9x2 + 30 xy + 25y2 (sifat 3) 2 (7x2 - 2xy)2 = (7x2)2 – 2 (7x2)( 2xy) + (2xy)2 = 49x2 – 28 x3y + 4x2y2 (sifat 4) 3 (x + y + 3)(x + y - 3) = (x + y)2 – 32 = x2 + 2xy + y2 -9 (2x – y – 1)(2x – y + 1) = (2x – y )2 – 12 = 4x2 – 4xy + y2 – 1 4 5 ( xy – 2)3 = (xy)3 – 3(xy)2.2 + 3 xy. 22 - 23 = x3y3 – 6x2y2 + 12 xy – 8 6 (x – 1)( x2 + x + 1) = x3 -1 (2x + 3y + z)2 = (2x)2 + (3y)2 + z2 + 2 (2x)(3y) + 2 ( 2x)z + 2(3y).z = 4x2 + 9y2 + z2 + 12 xy + 4 xz + 6 yz 7 Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  6. 8 ( x + y + z + 1)2 = [(x + y) + ( z + 1)]2 = (x + y)2 + 2(x + y) ( z + 1) + ( z + 1)2 = x2 + 2xy + y2 + 2(xz + x + yz + y) + z2 + 2z + 1 = x2 + 2xy + y2 + 2xz + 2x + 2yz + 2y + z2 + 2z + 1 9 (u – v)3 (u + v)3 = = {(u – v)(u + v)}3 = ( u2 – v2)3 = (u2)3 - 3 (u2)2 v2+ 3 u2(v2)2 –(v2)3 = u6 – 3 u4v2 + 3u2v4 – v6 10 (x2 – x + 1)2( x2 + x + 1)2 =…. = {(x2 – x + 1)( x2 + x + 1)}2 = {(x2 + 1– x)( x2 + 1+ x)}2 = {(x2 + 1)2– x2)} 2 = (x4 + 2x2 + 1 – x2)2 = ( x4 + x2 + 1)2 = (x4)2 +( x2)2 + 12 + 2 x6 + 2x4 + 2x2 (x2 – x + 1)2( x2 + x + 1)2 = x8 + 2x6 + 3x4 + 2x2 +1 Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  7. Soal Latihan 1 I. Carilah tiap – tiap hasil kali berikut (5xy + 4)(5xy – 4) ( 3 – 2x2)2 (xy + 6)(xy - 4 ) (2t2 + s)(3t2 + 4s) (x2 + 4y)(2x2 y – y) (r + s – 1)( r + s +1) (x – 2y +z)(x - 2y – z) (ab2 – 2b)3 (t – 2)(t2 + 2t + 4) (s – 1)(s3 + s2 + s + 1) (1 + t2)(1 - t2 + t4 – t6) (3x + 2y)2(3x – 2y)2 (x2+ 2x +1)2(x2 - 2x + 1)2 (y -1)3(y + 1)3 (u + 2)(u – 2)(u2 + 4)(u4 + 16) Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  8. II. Faktorkanlah Bentuk Berikut 3x2y4 + 6x3y3 = 12s2t2 – 6s5t4 + 4s4t = 4y2 – 100 = x2y2 – 8xy +16 = 4x3y + 12x2y2 + 9xy3 = y3 + 27 = x3y3 – 8 = 8x4y – 64xy4 = Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

  9. III. Sederhanakanlah Pecahan Berikut 1 5 2 6 3 7 4 Matematika Terapan 1 Modul 1 Aljabar by Ketut Darma Teknik Mesin Politeknik Negeri Bali

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