Phân tích đa thức thành nhân tử :
4a2+1 ( a < 0 )
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HN
2
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 11 2021
\(=4\left[a^2-\left(b^2-4bc+4c^2\right)\right]\)
\(=4\left[a^2-\left(b-2c\right)^2\right]\)
\(=4\left(a-b+2c\right)\left(a+b-2c\right)\)
MH
14 tháng 11 2021
\(4a^2-4b^2+16bc-16c^2\)
\(=4a^2-\left(4b^2-16bc+16c^2\right)\)
\(=\left(2a\right)^2-\left[\left(2b\right)^2-2.2b.4c+\left(4c\right)^2\right]\)
\(=\left(2a\right)^2-\left(2b-4c\right)^2\)
\(=\left(2a+2b-4c\right)\left(2a-2b+4c\right)\)
\(=4\left(a+b-c\right)\left(a-b+c\right)\)
PT
1
CM
30 tháng 11 2017
d) (8a3 – 27b3) – 2a(4a2 – 9b2)
= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)
= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)
2 tháng 10 2021
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
2 tháng 10 2021
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
15 tháng 7 2021
`a)x^4+2x^2y+y^2`
`=(x^2+y)^2`
`b)(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a)(2a+b+2b+a)`
`=(a-b)(3a+3b)`
`=3(a-b)(a+b)`
`c)8a^3-27b^3-2a(4a^2-9b^2)`
`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`
`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`
`=9b^2(2a-3b)`
16 tháng 7 2021
a) Ta có: \(x^4+2x^2y+y^2\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)
\(=\left(x^2+y\right)^2\)
b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
27 tháng 10 2021
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
M
1
6 tháng 1 2022
\(a,x^2-4xa+4a^2-81y^2=\left(x-2a\right)^2-\left(9y\right)^2=\left(x-2a-9y\right)\left(x-2a+9y\right)\\ b,3x^2-8x+4=\left(3x^2-6x\right)-\left(2x-4\right)=3x\left(x-2\right)-2\left(x-2\right)=\left(x-2\right)\left(3x-2\right)\)
22 tháng 11 2021
\(a,=\left(2a-1\right)\left(2a+1\right)\\ b,=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\\ c,=\left(16-3x\right)\left(16+3x\right)\\ d,Sửa:-36x^2+24x-4=-4\left(9x^2-6x+1\right)=-4\left(3x-1\right)^2\)
7 tháng 11 2021
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
7 tháng 11 2021
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
15 tháng 7 2021
a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)
b) \(a^2-9=\left(a-3\right)\left(a+3\right)\)
c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)
d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)
e) \(\left(2a+b\right)^2-a^2=\left(2a+b-a\right)\left(2a+b+a\right)=\left(a+b\right)\left(3a+b\right)\)
f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)
PD
15 tháng 7 2021
a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$
b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$
c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$
d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$
e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$
f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$
TN
1
23 tháng 11 2021
\(a,=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(b\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\\ b,a< 0\text{ nên biểu thức không phân tích đc}\)