把下列各式因式分解
10(a) 4x二次-九分之一y二次
10(b) –x六次+16x二次
10(c)8(x+y) 二次-2(x-y) 二次
10(d) x二次-8x+16
10(e) 4a二次+12ab+9b二次
10(f) 2x二次-2x+二分之一
10(g) (a-b)二次-10(a-b)c+25c二次
化簡下列各式
(a)(c二次+cd)-5(-c二次+cd+d二次)
(b)(x三次-x-7)+(2x二次+4x-9)
(c) (5n四次-2n三次-1)+(2+n二次-n三次)
(d)
a二次-a __________25
---------- 乘 ---
5___________a
(e) x __________ b二次+b
------- 乘 ------------
B___________ x二次y
(f) a二次b 2 ________ -2
----------- 乘 -----------
2-2a___________ ab二次
(g) 2 _________________1
---------- 乘 -----------
X(x+2) ___________ x(x+3)
www.bcaabc9999.somee.com/maths1.doc
(maths)把下列各式因式分解&化簡下列各式
2008-03-21 3:05 am
回答 (2)
2008-03-21 4:12 pm
✔ 最佳答案
10(a) 4x^2 - 1/9 * y^2= (2x+1/3y) (2x-1/3y)
10(b) –x^6+16x^2
= 16x^2 - x^6
= x^2 (16 - x^4)
= x^2 (4 - x^2) (4 + x^2)
= x^2 (2 - x) (2 + x) (4 + x^2)
2008-03-21 08:12:53 補充：
10(a) 4x^2 - 1/9 * y^2
= (2x+1/3y) (2x-1/3y)
10(b) –x^6+16x^2
= 16x^2 - x^6
= x^2 (16 - x^4)
= x^2 (4 - x^2) (4 + x^2)
= x^2 (2 - x) (2 + x) (4 + x^2)
10(c) 8(x+y)^2 - 2(x-y)^2
= 2 [ 4(x+y)^2 - (x-y)^2]
= 2 [ 2(x+y) - (x-y)][ 2(x+y) + (x-y)]
= 2 [ 2x+2y - x+y][ 2x+2y + x-y]
= 2 [ x+3y][ 3x+y]
10(d) x^2-8x+16
= (x-4)^2
10(e) 4a^2+12ab+9b^2
= (2a + 3b)^2
10(f) 2x^2-2x+1/2
= 1/2 (4x^2-4x+1)
= 1/2 (2x-1)^2
10(g) (a-b)^2-10(a-b)c+25c^2
= [(a-b) - 5c]^2
= [a-b- 5c]^2
(a) (c^2+cd)-5(-c^2+cd+d^2)
= c^2+cd+5c^2-5cd-5d^2
=6c^2-4cd-5d^2
(b) (x^3-x-7)+(2x^2+4x-9)
= x^3-x-7+2x^2+4x-9
= x^3+2x^2+3x-16
(c) (5n^4-2n^3-1)+(2+n^2-n^3)
=5n^4-2n^3-1+2+n^2-n^3
=5n^4-3n^3+n^2+1
2008-03-21 6:53 am
10(a) 4x^2 - 1/9 * y^2
= (2x+1/3y) (2x-1/3y)
10(b) –x^6+16x^2
= 16x^2 - x^6
= x^2 (16 - x^4)
= x^2 (4 - x^2) (4 + x^2)
= x^2 (2 - x) (2 + x) (4 + x^2)
10(c) 8(x+y)^2 - 2(x-y)^2
= 2 [ 4(x+y)^2 - (x-y)^2]
= 2 [ 2(x+y) - (x-y)][ 2(x+y) + (x-y)]
= 2 [ 2x+2y - x+y][ 2x+2y + x-y]
= 2 [ x+3y][ 3x+y]
10(d) x^2-8x+16
= (x-4)^2
10(e) 4a^2+12ab+9b^2
= (2a + 3b)^2
10(f) 2x^2-2x+1/2
= 1/2 (4x^2-4x+1)
= 1/2 (2x-1)^2
10(g) (a-b)^2-10(a-b)c+25c^2
= [(a-b) - 5c]^2
= [a-b- 5c]^2
(a) (c^2+cd)-5(-c^2+cd+d^2)
= c^2+cd+5c^2-5cd-5d^2
=6c^2-4cd-5d^2
(b) (x^3-x-7)+(2x^2+4x-9)
= x^3-x-7+2x^2+4x-9
= x^3+2x^2+3x-16
(c) (5n^4-2n^3-1)+(2+n^2-n^3)
=5n^4-2n^3-1+2+n^2-n^3
=5n^4-3n^3+n^2+1
For part (d) later on, I can't excatly get your meaning
= (2x+1/3y) (2x-1/3y)
10(b) –x^6+16x^2
= 16x^2 - x^6
= x^2 (16 - x^4)
= x^2 (4 - x^2) (4 + x^2)
= x^2 (2 - x) (2 + x) (4 + x^2)
10(c) 8(x+y)^2 - 2(x-y)^2
= 2 [ 4(x+y)^2 - (x-y)^2]
= 2 [ 2(x+y) - (x-y)][ 2(x+y) + (x-y)]
= 2 [ 2x+2y - x+y][ 2x+2y + x-y]
= 2 [ x+3y][ 3x+y]
10(d) x^2-8x+16
= (x-4)^2
10(e) 4a^2+12ab+9b^2
= (2a + 3b)^2
10(f) 2x^2-2x+1/2
= 1/2 (4x^2-4x+1)
= 1/2 (2x-1)^2
10(g) (a-b)^2-10(a-b)c+25c^2
= [(a-b) - 5c]^2
= [a-b- 5c]^2
(a) (c^2+cd)-5(-c^2+cd+d^2)
= c^2+cd+5c^2-5cd-5d^2
=6c^2-4cd-5d^2
(b) (x^3-x-7)+(2x^2+4x-9)
= x^3-x-7+2x^2+4x-9
= x^3+2x^2+3x-16
(c) (5n^4-2n^3-1)+(2+n^2-n^3)
=5n^4-2n^3-1+2+n^2-n^3
=5n^4-3n^3+n^2+1
For part (d) later on, I can't excatly get your meaning
參考: myself
收錄日期: 2021-04-16 14:02:35
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