If P + Q = 6 and PQ =3 ( P > Q), find P^5 + Q^5 =?
回答 (10)
2016-06-03 1:49 pm
P+Q=6.
P^2+Q^2= (P+Q)^2- 2PQ= 36-6=30.
Square it again
P^4 + Q^4 + 2 P^2Q^2=900,
P^4 + Q^4= 900- 2*9= 882.
(P^4 + Q^4) (P+Q)= P^5 + Q^5 + P^4Q+Q^4P= 882*6=5292
P^5+Q^5+PQ (P^3+Q^3)=5292
(P^2+Q^2)(P+Q)= 30*6=180= P^3+Q^3+PQ(P+Q)
P^3+Q^3= 180-3*6=162
P^5+Q^5= 5292-162*3=4806
P^2+Q^2= (P+Q)^2- 2PQ= 36-6=30.
Square it again
P^4 + Q^4 + 2 P^2Q^2=900,
P^4 + Q^4= 900- 2*9= 882.
(P^4 + Q^4) (P+Q)= P^5 + Q^5 + P^4Q+Q^4P= 882*6=5292
P^5+Q^5+PQ (P^3+Q^3)=5292
(P^2+Q^2)(P+Q)= 30*6=180= P^3+Q^3+PQ(P+Q)
P^3+Q^3= 180-3*6=162
P^5+Q^5= 5292-162*3=4806
2016-06-03 12:21 pm
(P + Q)^5 – (P^5 + Q^5) = 5PQ(P + Q) (P^2 + PQ + Q^2)
P^5 + Q^5 = (P + Q)^5 - 5PQ(P + Q)[(P + Q)^2 – PQ]
P^5 = 2403 + 981√(6)
6^5 - 15(6)[36 – 3] = 4806
Alternate: P = 3 + √(6) Q = 3 - √(6) or vice versa
P^5 = 2403 + 981√(6)
Q^5 = 2403 - 981√(6)
P^5 = 4806
P^5 + Q^5 = (P + Q)^5 - 5PQ(P + Q)[(P + Q)^2 – PQ]
P^5 = 2403 + 981√(6)
6^5 - 15(6)[36 – 3] = 4806
Alternate: P = 3 + √(6) Q = 3 - √(6) or vice versa
P^5 = 2403 + 981√(6)
Q^5 = 2403 - 981√(6)
P^5 = 4806
2016-06-03 11:48 am
Q = 3/P
P + 3/P = 6
Solve for P
Solve for Q
Finish your own math homework.
P + 3/P = 6
Solve for P
Solve for Q
Finish your own math homework.
2016-06-07 1:04 am
Don't you think that you are doing yourself a disservice by asking the question? You will benefit much more from working it out for yourself.
2016-06-03 3:11 pm
p+q = 6..[1]. pq = 3..[2]. (p+q)^5 = p^5 + q^5 + 5(p^4q + pq^4) + 10(p^3q^2 + p^2q^3) = p^5 + q^5 +
5pq(p^3 + q^3) + 10p^2q^2(p + q) = p^5 + q^5 + 5pq(p+q)(p^2 + q^2 - pq) + 10(p+q)p^2q^2 =
p^5 + q^5 + 5pq(p+q)[p^2 + q^2 - pq + 2pq] = p^5 + q^5 + 5*3*6[(p+q)^2 -pq] = p^5 + q^5 + 90(33).
Then p^5 + q^5 = (p+q)^5 - 90*33 = 6^5 - 90*33 = 4806.
5pq(p^3 + q^3) + 10p^2q^2(p + q) = p^5 + q^5 + 5pq(p+q)(p^2 + q^2 - pq) + 10(p+q)p^2q^2 =
p^5 + q^5 + 5pq(p+q)[p^2 + q^2 - pq + 2pq] = p^5 + q^5 + 5*3*6[(p+q)^2 -pq] = p^5 + q^5 + 90(33).
Then p^5 + q^5 = (p+q)^5 - 90*33 = 6^5 - 90*33 = 4806.
2016-06-03 1:55 pm
(P+Q)^5 =5P^4Q+10P^3Q^2+10P^2Q^3+5PQ^4
→P^5+Q^5=(P+Q)^5 −(5P^4Q+10P^3Q^2+10P^2Q^3+5PQ^4)
→P^5+Q^5=(P+Q)^5 −5PQ(P^3+2P^2Q+2PQ^2+Q^3)
→P^5+Q^5=(P+Q)^5 −5PQ(P^3+3P^2Q−P^2Q+3PQ^2−PQ^2+Q^3)
→P^5+Q^5=(P+Q)^5 −5PQ(P^3+3P^2Q+3PQ^2+Q^3−P^2Q−PQ^2)
→P^5+Q^5=(P+Q)^5 −5PQ{(P^3+3P^2Q+3PQ^2+Q^3)−PQ(P+Q)}
→P^5+Q^5=(P+Q)^5 −5PQ{(P+Q)^3−PQ(P+Q)}
→P^5+Q^5=(6)^5 −5×3{(6)^3−3(6)}
→P^5+Q^5=7776 −5×3{216−18}
→P^5+Q^5=7776 −15{198}
→P^5+Q^5=7776 −2970
→P^5+Q^5=4806
→P^5+Q^5=(P+Q)^5 −(5P^4Q+10P^3Q^2+10P^2Q^3+5PQ^4)
→P^5+Q^5=(P+Q)^5 −5PQ(P^3+2P^2Q+2PQ^2+Q^3)
→P^5+Q^5=(P+Q)^5 −5PQ(P^3+3P^2Q−P^2Q+3PQ^2−PQ^2+Q^3)
→P^5+Q^5=(P+Q)^5 −5PQ(P^3+3P^2Q+3PQ^2+Q^3−P^2Q−PQ^2)
→P^5+Q^5=(P+Q)^5 −5PQ{(P^3+3P^2Q+3PQ^2+Q^3)−PQ(P+Q)}
→P^5+Q^5=(P+Q)^5 −5PQ{(P+Q)^3−PQ(P+Q)}
→P^5+Q^5=(6)^5 −5×3{(6)^3−3(6)}
→P^5+Q^5=7776 −5×3{216−18}
→P^5+Q^5=7776 −15{198}
→P^5+Q^5=7776 −2970
→P^5+Q^5=4806
2016-06-03 12:28 pm
PQ = 3
P= 3/Q
P+Q = 6
P=6-Q
P=3/Q
6-Q = 3/Q
multiply both sides by Q
6Q-Q^2 = 3
-Q^2+6Q -3=0
Q^2-6Q+3 = 0
This equation is of form ax^2+bx+c=0
a = 1 b = -6 c = 3
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[6 +/-sqrt(-6^2-4(1)(3)]/(2)(1)
discriminant is b^2-4ac =24
x=[6 +√(24)] / (2)(1)
x=[6 -√(24)] / (2)(1)
x=[6+4.898979485566356] / 2
x=[6-4.898979485566356] / 2
The roots are 5.4495 and 0.5505
Q=0.5505
P=6-Q = 6 - 0.5505 = 5.4495
P > Q
P^5 + Q^5 = (5.4495)^5 +(0.5505)^5 = 4806.0452
P= 3/Q
P+Q = 6
P=6-Q
P=3/Q
6-Q = 3/Q
multiply both sides by Q
6Q-Q^2 = 3
-Q^2+6Q -3=0
Q^2-6Q+3 = 0
This equation is of form ax^2+bx+c=0
a = 1 b = -6 c = 3
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[6 +/-sqrt(-6^2-4(1)(3)]/(2)(1)
discriminant is b^2-4ac =24
x=[6 +√(24)] / (2)(1)
x=[6 -√(24)] / (2)(1)
x=[6+4.898979485566356] / 2
x=[6-4.898979485566356] / 2
The roots are 5.4495 and 0.5505
Q=0.5505
P=6-Q = 6 - 0.5505 = 5.4495
P > Q
P^5 + Q^5 = (5.4495)^5 +(0.5505)^5 = 4806.0452
2016-11-12 1:26 pm
this is nice asking
2016-06-05 6:10 am
P+3/P=3, P²-3P-3=0, P= (3±√9+12)/2 =
2016-06-03 11:33 am
69
收錄日期: 2021-05-01 20:48:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160603033148AAZY9QT