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a/ 9x^2 - 6x - y^2 + 1
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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.
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)
\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
16 tháng 11 2021
a) \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)
b) \(6x^2+x-2=2x\left(3x+2\right)-1\left(3x+2\right)=\left(3x+2\right)\left(2x-1\right)\)
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: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
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. \(6x^2-24x=0\)
\(\Leftrightarrow6x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
1 tháng 11 2023
a: \(9x^3y^2+3x^2y^2\)
\(=3x^2y^2\cdot3x+3x^2y^2\cdot1\)
\(=3x^2y^2\left(3x+1\right)\)
b: \(x^2-2x+1-y^2\)
\(=\left(x^2-2x+1\right)-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1-y\right)\left(x-1+y\right)\)
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: \(4xy-4x^2-y^2+9\)
\(=-\left(4x^2-4xy+y^2-9\right)\)
\(=-\left(2x-y-3\right)\left(2x-y+3\right)\)
NL
2
21 tháng 8 2021
\(-9x^2+6x+y^2-1\)
\(=-\left(9x^2-6x+1-y^2\right)\)
\(=-\left(3x-1-y\right)\left(3x-1+y\right)\)
NH
1
31 tháng 7 2021
a) \(9-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
b) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+y+3\right)\left(x-y+3\right)\)
GH
30 tháng 7 2023
a
\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)
b
\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)
30 tháng 7 2023
a: 8x^3-1/125y^3
=(2x)^3-(1/5y)^3
=(2x-1/5y)(4x^2+2/5xy+1/25y^2)
b: =(2y-x)^3
\(9x^2-6x-y^2+1\)
\(=\left(9x^2-6x+1\right)-y^2\)
\(=\left(3x-1\right)^2-y^2\)
\(=\left(3x-1-y\right)\left(3x-1+y\right)\)