中二數學....恆等式

引人入性，首兩條你的確不是直接計算，但你這樣只算是速算，而不是利用恆等式。

利用恆等式應是這樣計算﹕利用x²-y²=(x+y)(x-y)
1.
398² -397²
=(398+397)(398-397)
=(795)(1)
=795
2.
473²-472²
=(473+472)(473-472)
=(945)(1)
=945

1.
9a³-30a² b+25ab²
=a(9a²-30ab+25b²)
=a[(3a)²-2(3a)(5b)+(5b)²]
=a(3a-5b)²
2.
18a³-84a²b+98ab²
=2a(9a²-42ab+49b²)
=2a[(3a)²-2(3a)(7b)+(7b)²]
=2a(3a-7b)²

1.398² -397²
=(398+397)(398-397)
=795

2.473²-472²
=(473+472)(473-472)
=945

因式分解下列數式

1.9a^3-30a² b+25ab²
=(3a根號a-5根號a b)^2

2.18a^3-84a²b+98ab²
=2(9a^3-42a^2b+49ab^2)
=2( 3a根號a-7根號a b)^2

之後沒得再計了，用因式分解就這樣了。

1. 398² -397²
=(398+397)*(398-397)
=795*1
=795
2. 473²-472²
=(473+472)*(473-472)
=945*1
=945

1.9a^3-30a² b+25ab²
=a*(9a² -30ab+25b²)
=(3a-5b)²a

2.18a^3-84a²b+98ab²
=2a*(9a² -42ab+49b²)
=2(3a-7b)²a

注:step1:抽common factor
step2:有square的開方

1.398² -397²
by the identity a^2-b^2=(a+b)(a-b)
∴398² -397²
=(398+397)(398-397)
=795

2.473²-472²
=(473+472)(473-472)
=945

因式分解下列數式(但要利用恆等式(x-y)²=x²-2(x)(y)+y²):

1.9a^3-30a² b+25ab²
=a(9a^2-30ab+25b^2)
=a(3a-5b)^2

2.18a^3-84a²b+98ab²
=2a(9a^2-42ab+49b^2)
=2a(3a-7b)^2

記得學好ｄ因式分解呀
你考會考會有好多野都用到因式分解架

1.398² -397²
= (398-0)^2 - (398-1)^2
= (398^2 -0 +0) - (398^2 - 2x398 +1)
= 2x398 -1
= 795

2.473²-472²
= (473-0)^2 - (473-1)^2
= (473^2 -0 +0) - (473^2 - 2x473 +1)
= 2x473 -1
= 945

1.9a^3-30a² b+25ab²
= a[(3a)^2 - 2(3a)(5b) +(5b)^2]
= a(3a-5b)^2

2.18a^3-84a²b+98ab²
= 2a[(3a)^2 - 2(3a)(7b) +(7b)^2]
= 2a{3a-7b)^2

2007-10-19 21:36:52 補充：
(398-0)^2 = [398^2 - 2(398)(0) +0^2] = (398^2 -0 +0)以上係用0左恆等式(x-y)²=x²-2(x)(y)+y²其它 (398-1)^2 , (473-0)^2 同 (473-1)^2 都係一樣