Phân tích thành nhân tử
a)\(6x^2+23x-18\)
b)\(x^4+324\)
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16 tháng 11 2021
a) \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)
b) \(6x^2+x-2=2x\left(3x+2\right)-1\left(3x+2\right)=\left(3x+2\right)\left(2x-1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 1 2024
a.
\(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b.
\(x^3-9x^2+6x+16=\left(x^3-7x^2-8x\right)-\left(2x^2-14x-16\right)\)
\(=x\left(x^2-7x-8\right)-2\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)=\left(x-2\right)\left(x^2+x-8x-8\right)\)
\(=\left(x-2\right)\left[x\left(x+1\right)-8\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x-8\right)\)
c.
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)
\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
11 tháng 11 2021
2.
a) 4x(x-1)-6x+6
= 4x(x-1)-6(x-1)
= (4x-6)(x-1)
3.
a) 6x2-24x=0
6x(x-4)=0
TH1: 6x=0 TH2: x-4=0
x=0 x=4
Vậy x\(\in\){0;4}
HP
11 tháng 11 2021
2. a. \(4x\left(x-1\right)-6x+6\)
\(=4x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(4x-6\right)\left(x-1\right)\)
3. a. \(6x^2-24x=0\)
\(\Leftrightarrow6x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
11 tháng 8 2021
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
11 tháng 8 2021
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
15 tháng 8 2021
a: Ta có: \(x^2-6x+9-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-y-3\right)\left(x+y-3\right)\)
b: Ta có: \(x^3+4x^2+4x\)
\(=x\left(x^2+4x+4\right)\)
\(=x\left(x+2\right)^2\)
c: Ta có: \(4xy-4x^2-y^2+9\)
\(=-\left(4x^2-4xy+y^2-9\right)\)
\(=-\left(2x-y-3\right)\left(2x-y+3\right)\)
TY
0
23 tháng 11 2021
\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)
23 tháng 11 2021
a) 2x2+6x=2x(x+3)
b) x4+3x3+x+3=(x4+x)+(3x3+3)=x(x3+1)+3(x3+1)=(x+3)(x3+1)
c) 64-x2-y2+2xy=-(x2-2xy+y2)+82=8-(x+y)2=(8+x+y)(8-x-y)
A= (x+5)(x+1)+(x-2)(x2+2xx+4)-(x2+x-2)
A= x2+6x+5+x3-8-x2-x+2
A= x3+(x2-x2)+(6x-x)+(5-8+2)
A= x3+5x-1
a)6x2+23x-18
=6x2-4x+27x-18
=2x(3x-2)+9(3x-2)
=(2x+9)(3x-2)
b)x4+324
=(x4+182+36x2)-36x2
=(x2+18)2-36x2
=(x2-6x+18)(x2+6x+18)
a,6x2+23x-18
=6x2+27x-4x-18
=(6x2+27x)-(4x+18)
=3x(2x+9)-2(2x-9)
=(3x-2)(2x+9)