bài 2: Phân tích các đa thức sau thành nhân tử
a. 3x^2+5y-3xy-5x
b. 3y^2-3z^2+3x^2+6xy
c.x^2-25-2xy+y^2
d.5x^2-10xy+5y^2-20z^2
e. x^2-5x+5y-y^2
f. 3x^2-6xy+3y^2-12z^2
K
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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)
TA
17 tháng 12 2017
a.5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[ (x2-2xy+y2)-(2z)2 ]
=5[ (x-y)2-(2z)2 ]
=5(x-y-2z)(x-y+2z)
b.16x-5x2-3
=15x+x-5x2-3
=(15x-3)+(x-5x2)
=3(5x-1)+x(1-5x)
=3(5x-1)-x(5x-1)
=(5x-1)(3-x)
c.x2-5x+5y-y2
=(5y-5x)+(x2-y2)
=5(y-x)+(x-y)(x+y)
=5(y-x)-(y-x)(y+x)
=(y-x)[5-(y+x)]
=(y-x)(5-y-x)
d.3x2-6xy+3y2-12z2 (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)
=3(x2-2xy+y2-4z2)
=3[ (x2-2xy+y2)-(2z)2 ]
=3[ (x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
e.x2+4x+3
=x2+3x+x+3
=(x2+x)+(3x+3)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f.(x2+1)2-4x2
=(x2+1)2-(2x)2
=(x2+1-2x)(x2+1+2x)
h.x2-4x-5
=x2-5x+x-5
=(x2+x)+(-5x-5)
=x(x+1)-5(x+1)
-(x+1)(x-5)
AM
19 tháng 6 2015
b)x2+2xy+y2-16=(x+y)2-42=(x+y+4)(x+y-4)
c)3x2+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)
d)4x2-6x3y-2x2+8x=2x(2x-3x2y-x+4)
e)x2-4-2xy+y2=(x2-2xy+y2)-4=(x-y)2-22=(x-y-2)(x-y+2)
k)x2-y2-z2-2yz=x2-(y+z)2=(x-y-z)(x+y+z)
m)6xy+5x-5y-3x2-3y2=3(x2-2xy+y2)+5(x-y)=3(x-y)2+5(x-y)=(x-y)(3x-3y+5)
NT
2
NC
5 tháng 7 2021
a.\(xz+yz-5\left(x+y\right)\)
\(=z\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(z-5\right)\)
b.\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?
d.\(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)'
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
QA
5 tháng 7 2021
Trả lời:
a, xz + yz - 5 ( x + y )
= ( xz + yz ) - 5 ( x + y )
= z ( x + y ) - 5 ( x + y )
= ( x + y ) ( z - 5 )
b, 3x2 - 3xy - 5x + 5y
= ( 3x2 - 3xy ) - ( 5x - 5y )
= 3x ( x - y ) - 5 ( x - y )
= ( x - y ) ( 3x - 5 )
c, x2 + 6x - y2 - 3z2
= - ( 3x2 - x2 + y2 - 6x )
d, 3x2 + 6xy + 3y2 - 3z2
= 3 ( x2 + 2xy + y2 - x2 )
= 3 [ ( x2 + 2xy + y2 ) - z2 ]
= 3 [ ( x + y )2 - z2 ]
= 3 ( x + y - z ) ( x + y + z )
NH
1
11 tháng 12 2018
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
2 tháng 12 2018
a) \(3x^2-6x+9x^2\)
\(=12x^2-6x\)
\(=6x\left(2x-1\right)\)
b) hỏi chấm đề ?
c) \(3x^2+5y-3xy-5x\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
d) \(3y^2-3z^2+3x^2-6xy\)
\(=3\left(y^2-z^2+x^2-2xy\right)\)
\(=3\left[\left(x-y\right)^2-z^2\right]\)
\(=3\left(x-y-z\right)\left(x-y+z\right)\)
e) \(16x^3+54y^3\)
\(=2\left(8x^3+27y^3\right)\)
\(=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
f) \(x^2-25-2xy+y^2\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
g) \(x^5-3x^4-3x^3-x^2\)
\(=x^2\left(x^3-3x^2-3x-1\right)\)
H
16 tháng 12 2018
\(10x\left(x-y\right)-6y\left(y-x\right)\)
\(=10x\left(x-y\right)+6x\left(x-y\right)\)
\(=\left(10x+6x\right)\left(x-y\right)\)
\(c,3x^2+5y-3xy-5x\)
\(=\left(3x^2-3xy\right)+\left(5y-5x\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(3x-5\right)\left(x-y\right)\)
\(e,27+27x+9x^2=3\left(9+9x+x^2\right)\)
a) Ta có: \(3x^2+5y-3xy-5x\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) Ta có: \(3y^2-3z^2+3x^2+6xy\)
\(=3\left(y^2-z^2+x^2+2xy\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
c) Ta có: \(x^2-25-2xy+y^2\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
d) Ta có: \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
e) Ta có: \(x^2-5x+5y-y^2\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f) Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)