Phân tích đa thức thành nhân tử:
a)9(a+b)^2-4(a-2b)^2
b)9x^6-12x^7+4x^8
c)8x^6-27y^3
d)1/64x^6-125y^3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
16 tháng 10 2016
a,
=\(\left(a^2\right)^2-\left(2b\right)^2\)
=\(\left(a^2-2b\right)\left(a^2+2b\right)\)
= \(\left(\left(a-\sqrt{2b}\right)\left(a+\sqrt{2b}\right)\right)\left(a^2+2b\right)\)
c,
=\(4x^4+20x^2+25\)
=\(\left(2x^2\right)^2+2.2x^2.5+5^2\)
=\(\left(2x^2+5\right)^2\)
d,
=\(8x^6-27y^3\)
= \(\left(2x^2\right)^3-\left(3y\right)^3\)
= \(\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
Câu b đề ghi ko rõ lắm
12 tháng 7 2023
a: =xy(x^2-4xy^2+4y^4)
=xy(x-2y^2)^2
b:=(x^3-y)^2
c: =(a^2-b^2)(a^2+b^2)
=(a^2+b^2)(a-b)(a+b)
d: 64x^6-27y^6
=(4x^2-3y^2)(16x^4+12x^2y^2+9y^4)
e: =(2x)^3+(3y)^3
=(2x+3y)(4x^2-6xy+9y^2)
16 tháng 8 2016
\(9x^4-12x^2y^3+16y^6=\left(3x^2+4y^3\right)^2\)
\(4x^4-16x^2y^3+16y^6=\left(2x^2-4y^3\right)^2\)
\(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
\(a^3+\frac{1}{27}=\left(a+\frac{1}{3}\right)\left(a^2-\frac{1}{3}a+\frac{1}{9}\right)\)
13 tháng 8 2023
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
LN
19 tháng 8 2021
\(a.6x^3-9x^2=3x^2\left(2x-3\right)\\ b.20x^2y-12x^3=4x^2\left(5y-3x\right)\)
19 tháng 8 2021
a) \(6x^3-9x^2=3x^2\left(2x-3\right)\)
b) \(20x^2y-12x^3=4x^2\left(5y-3x\right)\)
30 tháng 7 2020
1: Ta có: \(x^{10}-4x^8+4x^6\)
\(=x^6\left(x^4-4x^2+4\right)\)
\(=x^6\left(x-2\right)^2\left(x+2\right)^2\)
2: Ta có: \(m^3+27\)
\(=\left(m+3\right)\left(m^2-3m+9\right)\)
3: Ta có: \(x^3+8\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
4: Ta có: \(\frac{1}{27}+a^3\)
\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)
5: Ta có: \(8x^3+27y^3\)
\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
6: Ta có: \(\frac{1}{8}x^3+8y^3\)
\(=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)
7: Ta có: \(8x^6-27y^3\)
\(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
8: Ta có: \(\frac{1}{8}x^3-8\)
\(=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)
9: Ta có: \(\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)
10: Ta có: \(\left(a+b\right)^3-c^3\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)
\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)
11: Ta có: \(x^3-\left(y-1\right)^3\)
\(=\left[x-\left(y-1\right)\right]\cdot\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]\)
\(=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)
12: Ta có: \(x^6+1\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
TG
30 tháng 7 2020
1) \(x^{10}-4x^8+4x^6\)
\(=x^6\left(x^4-4x^2+4\right)\)
2) \(m^3+27=m^3+3^3=\left(m+3\right)\left(m^2-3m+3^2\right)\)
3) \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+2^2\right)\)
4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)
5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)
8) \(\frac{1}{8}x^3-8=\left(\frac{1}{2}x\right)^3-2^3=\left(\frac{1}{2}x-2\right)\left[\left(\frac{1}{2}x\right)^2+\frac{1}{2}x.2+2^2\right]=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)
10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)
P/s: Đăng ít thôi chớ bạn!
a) \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)
\(=\left(3a+3b\right)^2-\left(2a-4b\right)^2\)
\(=\left(3a+3b-2a+4b\right)\left(3a+3b+2a-4b\right)\)
\(=\left(a+7b\right)\left(5a-b\right)\)
b) \(9x^6-12x^7+4x^8\)
\(=x^6\left(9-12x+4x^2\right)\)
\(=x^6\left(2x-3\right)^2\)
c) \(8x^6-27y^3\)
\(=\left(2x^2\right)^3-\left(3y\right)^3\)
\(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
d) \(\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)
\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{6}xy+25y^2\right)\)