Factorize each of the following
1. 144r² – 96r + 16q²
2. -s² - 2st - t²
3. 18x³ + 60x²yz + 50xy²z²
F. 2 Maths 3題
2007-04-21 9:48 pm
回答 (4)
2007-04-21 9:53 pm
✔ 最佳答案
1. 144r² – 96r + 16
= (12r + 4)²
2.
-s² - 2st - t²
= -(s²+ 2st + t²)
= -(s+t)²
3.
18x³ + 60x²yz + 50xy²z²
= 2x(9x² + 30xyz + 25y²z²)
= 2x(9x² + 30x(yz) + 25(yz)²)
= 2x(3x + 5yz)²
2007-04-21 15:05:15 補充：
sorry for the typothe 1st one should be (12r - 4)^2which can in turn be written as 16(3r - 1)^2
2007-04-21 10:18 pm
1. 144r^2 - 96r +16
=(12r)^2 -96r + 4^2
By identity of (x+y)^2=x^2+2xy+y^2,(12r)^2+2(12)(4)+4^2
=(12r+4)^2
2. -s^2-2st-t^2
=-(s^2+2st+t^2)
By identity of (x+y)^2=x^2+2xy+y^2,-(s^2+2st+t^2)
=-(s+t)^2
3. 18x^3+60x^2yz+50xy^2z^2
=2x(9x^2+30xyz+25y^2z^2)
By identity of (x+y)^2=x^2+2xy+y^2,2x[(3x)^2+2(3x)(5yz)+(5yz)^2)]
=(3x+5yz)^2
=(12r)^2 -96r + 4^2
By identity of (x+y)^2=x^2+2xy+y^2,(12r)^2+2(12)(4)+4^2
=(12r+4)^2
2. -s^2-2st-t^2
=-(s^2+2st+t^2)
By identity of (x+y)^2=x^2+2xy+y^2,-(s^2+2st+t^2)
=-(s+t)^2
3. 18x^3+60x^2yz+50xy^2z^2
=2x(9x^2+30xyz+25y^2z^2)
By identity of (x+y)^2=x^2+2xy+y^2,2x[(3x)^2+2(3x)(5yz)+(5yz)^2)]
=(3x+5yz)^2
參考: myself ...I hope I can help you la~
2007-04-21 10:14 pm
1. 144r² – 96r + 16q²
= (12r)² -2(4)(q)+(4q)²
= (12r - 4q)²
2. -s² - 2st - t²
= -(s² + 2st + t²)
=-[(s)² + 2(s)(t) + (t)²]
=-(s+t)²
= (12r)² -2(4)(q)+(4q)²
= (12r - 4q)²
2. -s² - 2st - t²
= -(s² + 2st + t²)
=-[(s)² + 2(s)(t) + (t)²]
=-(s+t)²
參考: me
2007-04-21 10:03 pm
Q1:
144r^2-96r+16
= (12r-4)(12r-4)
= (12r-4)^2
Q2:
-s^2-2st-t^2
= (s+t)(-s-t)
Q3:
18x^3+60x^2yz+50xy^2z^2
= x(18x^2+60xyz+50y^2z^2)
= x[(3x+5yz)(6x+10yz)]
= x(3x+5yz)(6x+10yz)
144r^2-96r+16
= (12r-4)(12r-4)
= (12r-4)^2
Q2:
-s^2-2st-t^2
= (s+t)(-s-t)
Q3:
18x^3+60x^2yz+50xy^2z^2
= x(18x^2+60xyz+50y^2z^2)
= x[(3x+5yz)(6x+10yz)]
= x(3x+5yz)(6x+10yz)
收錄日期: 2021-04-13 13:46:14
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https://hk.answers.yahoo.com/question/index?qid=20070421000051KK02227