إذا كان c(x) = x2 – 2x , d(x) = 3x2 – 6x + 4 ، فأوجد قيمة كل مما يأتى
c(a – 3) + d(a + 1)
الاجابة
c(a – 3) = (a – 3)2 – 2(a – 3)
c(a – 3) = (a2 – 6a + 9) – 2(a – 3)
c(a – 3) = a2 – 6a + 9 – 2a + 6
c(a – 3) = a2 – 8a + 15
d(a + 1) = 3(a + 1)2 – 6(a + 1) + 4
d(a + 1) = 3(a2 + 2a + 1) – 6(a + 1) + 4
d(a + 1) = 3a2 + 6a + 3 – 6a- 6 + 4
d(a + 1) = 3a2 + 1
c(a – 3) + d(a + 1) = a2 – 8a + 15 + 3a2 + 1
c(a – 3) + d(a + 1) = 4a2 – 8a + 16