Pt đa thức thành nhân tử
Câu a : 3x-3+5×(1-x)
Câu b :12a^2 -3ab+8ac-2bc
Câu c : x^2-25+y^2-2xy
K
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UT
3
5 tháng 1 2022
Bài 2:
a: =>4x(x+5)=0
=>x=0 hoặc x=-5
b: =>(x+3)(x-3)=0
=>x=-3 hoặc x=3
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
SO
2
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2021
a.
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)
b.
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c.
\(=x^4-1+4x^2-4\)
\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
16 tháng 7 2021
a) Ta có: \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
b) Ta có: \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
MH
3 tháng 10 2021
1) \(2xy^3-6x^2+10xy\)
\(=2x.y^3-2x.3x+2x.5y\)
\(=2x\left(y^3-3x+5y\right)\)
\(=2x[y\left(y^2-5\right)-3x]\)
T
13 tháng 8 2023
1)
a) => 16x2 - 8x + 1 - 8(2x2 + 3x - 4x - 6) = 15
=> 16x2 - 8x + 1 - 8(2x2 - x - 6) = 15
=> 16x2 - 8x + 1 - 16x2 + 8x + 48 = 15
=> 49 = 15 (?) (vô lí)
=> Không tìm được x thoả mãn
b) (5x - 2)(x - 2) - 4(x - 3) = x2 + 3
=> 5x2 - 10x - 2x + 4 - 4x + 12 = x2 + 3
=> 5x2 - 16x + 16 = x2 + 3
=> 4x2 - 16x + 16 = 3
=> (2x)2 - 2.2x.4 + 42 = 3
=> (2x - 4)2 = 3
=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)
Mong bạn xem lại đề bài!
T
13 tháng 8 2023
2)
a) 5x2 - 10xy + 5y2 - 20z2
= 5(x2 - 2xy + y2 - 4z2)
= 5[(x - y)2 - (2z)2]
= 5(x - y - 2z)(x - y + 2z)
b) a3 - ay - a2x + xy
= a(a2 - y) - x(a2 - y)
= (a - x)(a2 - y)
c) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x - y)2 - (2z)2]
= 3(x - y - 2z)(x - y + 2z)
d) x2 - 2xy + tx - 2ty
= x(x - 2y) + t(x - 2y)
= (x + t)(x - 2y)
PT
3
28 tháng 8 2023
`#040911`
`a)`
`x^2 + y^2 + 2xy - 25`
`= (x^2 + 2xy + y^2) - 25`
`= [ (x)^2 + 2*x*y + (y)^2] - 5^2`
`= (x + y)^2 - 5^2`
`= (x + y - 5)(x + y + 5)`
`b)`
`x^2 + 2x - 15`
`= x^2 + 5x - 3x - 15`
`= (x^2 + 5x) - (3x + 15)`
`= x(x + 5) - 3(x + 5)`
`= (x - 3)(x + 5)`
`c)`
`x^2 - x - 2`
`= x^2 - 2x + x - 2`
`= (x^2 - 2x) + (x - 2)`
`= x(x - 2) + (x - 2)`
`= (x + 1)(x - 2)`
`d)`
`3x^2 - 11x + 6`
`= 3x^2 - 9x - 2x + 6`
`= (3x^2 - 9x) - (2x - 6)`
`= 3x(x - 3) - 2(x - 3)`
`= (3x - 2)(x - 3)`
28 tháng 8 2023
`a, (x+y)^2-25 = (x+y+5)(x+y-5)`.
`b, x^2+2x-15 = (x+1)^2-16 = (x-3)(x+5)`.
`c, x^2-x-2=(x-2)(x+1)`
`d, 3x^2-11x+6 = (3x-2)(x-3)`.
25 tháng 7 2021
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
25 tháng 7 2021
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
DD
Đoàn Đức Hà
Giáo viên
28 tháng 5 2022
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
a) \(3x-3+5\left(1-x\right)\)
\(=3x-3+5-5x\)
\(=3x-5x+2\)
\(=x\left(3-5\right)+2\)
\(=-2x+2\)
\(=2\left(1-x\right)\)
b) \(12a^2-3ab+8ac-2bc\)
\(=3a\left(4a-b\right)+2c\left(4a-b\right)\)
\(=\left(4a-b\right)\left(3a+2c\right)\)
c) \(x^2-25+y^2-2xy\)
\(=x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)