PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ
a)\(4x^3+4x^2+x\)
b) \(x^2-2xy+y^2-49\)
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NN
16 tháng 10 2021
a,x^2-x-y^2-y
=x^2-y^2-(x+y)
=(x-y).(x+y)-(x+y)
=(x+y).(x-y-1)
b, x^2-2xy+y^2-z^2
=(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
c,5x-5y+ax-ay( đề bài ở đây phải là -ay ms tính đc)
=(5x-5y)+(ax-ay)
=5(x-y)+a(x-y)
=(x-y).(5+a)
d,a^3-a^2.x-ay+xy
=(a^3-a^2x)-(ay-xy)
=a^2(a-x)-y(a-x)
=(a-x)(a^2-y)
e,4x^2-y^2+4x+1
={(2x)^2+4x+1}-y^2
=(2x+1)^2-y^2
=(2x+1+y^2)(2x+1-y^2)
f,x^3-x+y^3-y
=(x^3+y^3)-(x+y)
=(x+y)(x^2-xy+y^2)-(x+y)
=(x+y)(x^2-xy+y^2-1)
14 tháng 10 2021
c: \(x^2-4+3\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(3x-6\right)\)
\(=\left(x-2\right)\left(x+2+3x-6\right)\)
\(=\left(4x-4\right)\left(x-2\right)\)
\(=4\left(x-1\right)\left(x-2\right)\)
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: \(4xy-4x^2-y^2+9\)
\(=-\left(4x^2-4xy+y^2-9\right)\)
\(=-\left(2x-y-3\right)\left(2x-y+3\right)\)
NM
1
4 tháng 9 2021
a) \(4x^2\left(x+3\right)-8x\left(3+x\right)=4x\left(x+3\right)\left(x-2\right)\)
b) \(4x^2+y^2-25+4xy=\left(2x+y\right)^2-25=\left(2x+y-5\right)\left(2x+y+5\right)\)
c) \(\left(x-3\right)^2-\left(x+2\right)^2=\left(x-3-x-2\right)\left(x-3+x+2\right)=-5\left(2x-1\right)\)
MH
1 tháng 11 2021
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
KJ
1 tháng 11 2021
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
TL
20 tháng 8 2021
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
KV
10 tháng 12 2023
a) x² + 6x + 8
= x² + 2x + 4x + 8
= (x² + 2x) + (4x + 8)
= x(x + 2) + 4(x + 8)
= (x + 2)(x + 4)
b) 3x² - 2(x - y)² - 3y²
= (3x² - 3y²) - 2(x - y)²
= 3(x² - y²) - 2(x - y)²
= 3(x + y)(x - y) - 2(x - y)²
= (x - y)[3(x + y) - 2(x - y)]
= (x - y)(3x + 3y - 2x + 2y)
= (x - y)(x + 5y)
c) 4x² - 9y² + 4x - 6y
= (4x² - 9y²) + (4x - 6y)
= (2x - 3y)(2x + 3y) + 2(2x - 3y)
= (2x - 3y)(2x + 3y + 2)
d) x(x + 1)² + x(x - 5) - 5(x + 1)²
= [x(x + 1)² - 5(x + 1)²] + x(x - 5)
= (x + 1)²(x - 5) + x(x - 5)
= (x - 5)[(x + 1)² + x]
= (x - 5)(x² + 2x + 1 + x)
= (x - 5)(x² + 3x + 1)
e) 2xy - x² + 3y² - 4y + 1
= -x² + 2xy - y² + 4y² - 4y + 1
= -(x² - 2xy + y²) + (4y² - 4y + 1)
= -(x - y)² + (2y - 1)²
= (2y - 1)² - (x - y)²
= (2y - 1 - x + y)(2y - 1 + x - y)
= (3y - x - 1)(x + y - 1)
f) 4x¹⁶ + 81
= (2x⁸)² + 2.2x⁸.9 + 9² - 2.2x⁸.9
= (2x⁸ + 9)² - 36x⁸
= (2x⁸ + 9) - (6x⁴)²
= (2x⁸ + 9 - 6x⁴)(2x⁸ + 9 + 6x⁴)
= (2x⁸ - 6x⁴ + 9)(2x⁸ + 6x⁴ + 9)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
a.
\(2x^3-x^2y+x^2+y^2-2xy-y=0\)
\(\Leftrightarrow x^2\left(2x-y+1\right)-y\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-y=0\\2x-y+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x^2\\y=2x+1\end{matrix}\right.\)
Thế vào pt đầu:
\(\left[{}\begin{matrix}x^3+x-2=0\\x\left(2x+1\right)+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2+x+2\right)=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
b.
\(x^2-2xy+x=-y\)
Thế vào \(y^2\) ở pt dưới:
\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)
\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)
\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)
\(\Leftrightarrow-2y+4y^2-8y+4=0\)
\(\Leftrightarrow...\)
a: =x(4x^2+4x+1)
=x(2x+1)^2
b: =(x-y)^2-49
=(x-y-7)(x-y+7)