solve the following inequality 2. SOLVE x∈R a) x+3/x-4>0 b) x-7/x-2 ≥0
Answers
Step-by-step explanation:
ANSWER
(x−2)(x−3)
7
+
x−3
9
+1<0
(x−2)(x−3)
7+9(x−2)+x
2
−5x+6
<0
(x−2)(x−3)
x
2
+4x−5
<0
(x−2)(x−3)
(x+5)(x−1)
<0
x∈(−5,1)∪(2,3)
Answer:
Problem 8. Prove that if x and y are real numbers, then 2xy ≤ x2 +y2. Proof. First we prove that if x is a real number, then x2 ≥ 0. The product of two positive numbers is always positive, i.e., if x ≥ 0 and y ≥ 0, then xy ≥ 0. In particular if x ≥ 0 then x2 = x·x ≥ 0. If x is negative, then −x is positive, hence (−x)2 ≥ 0. But we can conduct the following computation by the associativity and the commutativity of the product of real numbers: 0 ≥(−x)2 = (−x)(−x) = ((−1)x)((−1)x) = (((−1)x))(−1))x =(((−1)(x(−1)))x = (((−1)(−1))x)x = (1x)x = xx = x2. The above change in bracketting can be done in many ways. At any rate, this shows that the square of any real number is non-negaitive. Now if x and y are real numbers, then so is the difference, x − y which is defined to be x+(−y). Therefore we conclude that 0 ≤ (x+(−y))2 and compute: 0 ≤(x+(−y))2 = (x+(−y))(x+(−y)) = x(x+(−y))+(−y)(x+(−y)) =x2+x(−y)+(−y)x+(−y)2 = x2 +y2 +(−xy)+(−xy) =x2+y2+2(−xy); adding 2xy to the both sides, 2xy = 0+2xy ≤(x2+y2+2(−xy))+2xy = (x2 +y2)+(2(−xy)+2xy) =(x2 +y2)+0 = x2+y2. Therefore, we conclude the inequality: 2xy ≤ x2 +y2 for every pair of real numbers x and y. 1