Page 90 - 10. SINIF VIP TÜM DERSLER KONU ANLATIMLI - EDİTÖR YAYINLARI

90
̛ Örnek: Aşağıda verilen rasyonel ifadelerdeki çarpma ve bölme işlemlerini yapalım.

2
x +− ⋅ x1 x 2 +−x 6 ⋅ x1 = (x − + 2)(x 3) (x 1) = x2
+
x 6
+
+
−
⋅
−
+
+
+
−
x − 2 3x 4 x3 2 − x − 3x 4 x3 (x +1)(x 4) (x 3 ) x4
−
−
+
x − 2 2x 15 4x 12− : + x − 2 2x 15 4x 12 = (x 5) (x 3)+ ⋅ x1+ = 1
−
:
+
+
−
−
x − 2 4x 5 x1 x − 2 4x 5 x1 (x 5)− (x 1)+ 4(x 3)+ 4
6x + 2 5x 6− : 4x − 2 9 6x + 2 5x 6− : 4x − 2 9 = (3x 2)(2x 3)− + ⋅ (2x 3)− (2x 3)− = 3x 2−
4x − 2 6x 4x − 2 12x 9+ 4x − 2 6x 4x − 2 12x 9+ 2x (2x 3)− (2x 3)− (2x 3)+ 2x
3
2
x − 4 x 3 ⋅ x + 2 1 x (x 1)− ⋅ x + 1 = x 3 = x 3 = 1
EDİTÖR YAYINLARI
2
2
2
x + 4 x 2 x − 2 x x (x + 1) x (x 1)− x ⋅ x x 3
1 BENDEN 1 SENDEN TEST 9
x x A B 4x 5+
1. + − 4 3. + =
−
−
−
+
x 1 x1 x 1 x + 4 x + 2 3x 4
ifadesinin en sade hali aşağıdakilerden hangisi- olduğuna göre, A - B farkı kaçtır?
dir? 1 2 4 6
A) − B) − C) − D) - 1 E) −
2x− 2 4 2x− 2 4 5 5 5 5
A) 2 B) C) D) 4 E)
x − 1 x − 1 x − 1
2
2
2
A + B = + Ax +4A Bx B
−
Çözüm:
2
2
x
−
+
Çözüm: x + x + − 4 = x −+ x + x 4x− 2 + 4 x1 x 4 2 + x − 3x 4
−4)
2
−
+
x 1 x1 1 x − 1 (x + (x 1)
+
−
(x 1) (x 1) (x − 1) = + x(A B) (4A B) = 2 + x +4x 5 − 3x 4
2
−
+
2
+ x
− 3x 4
− 2x + 2 4 4 2x 2 + AB = 4
−
= =
x − 2 1 x − 2 1 + − 4A B = 5
9 11
= 5A ⇒ 9 = A = ve B
5 5
9 11 −2
A B − = olur.
−=
5 5 5
x + 3 y 3 x − 2 y 2
2. + 4. A B 7x 6− olduğuna göre A + B iş-
−
(x y ) + 2 xy x + y x 2 + x 3 = x −− 2B
2
+
x6
−
ifadesinin en sade hali aşağıdakilerden hangisi- leminin sonucu kaçtır?
dir? 1 5 6 3 7
A) B) C) D) E)
A) 0 B) -2x C) 2x D) x + y E) x - y 6 7 5 7 6
Çözüm:
Çözüm: