phân tích đa thức thành nhân tử:
\(5x^2-10xy+5y^2-20z^2\)
\(x^2\left(1-x^2\right)-4+4x^2\)
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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)
x2.(1 - x2) - 4 + 4x2
= x2.(1 - x2) - 4.(1 - x2)
= (1 - x2).(x2 - 4)
= (1 - x)(1 + x)(x - 2)(x + 2)
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)
5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[(x-y)2-4z2]
=5(x-y-2z)(x-y+2z)
dễ hiểu thôi.hơi vắn tắt
a) x4 + 2x3 + x2
= x2 ( x2 + 2x + 1 )
= x2 ( x + 1 )2
b) 5x2 - 10xy + 5y2 - 20z2
= 5 [(x2 - 2xy + y2 ) - 4z2 ]
= 5 [( x - y )2 - ( 2z )2 ]
= 5 ( x - y - 2z ) ( x - y + 2z )
c) x3 - x + 3x2y + 3xy2+ y3- y
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x + y )
= (x + y )3 - ( x + y)
= ( x + y ) [( x + y )2 - 1 ]
= ( x + y ) ( x + y + 1 ) ( x + y - 1 )
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)
5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[(x-y)2-(2z)2]
=5(x-y-2z)(x-y+2z)
a) \(5x^2+5xy-x-y\)
\(=5x.\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
b) \(5x^2-10y+5y^2-20z^2\)
\(=5.\left(x^2-2y+y^2-4z^2\right)\)
Đề sai ở đâu đó.
c) \(4x^2-y^2+4x+1\)
\(=\left(4x+4x^2+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+y+1\right)\left(2x-y+1\right)\)
a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)
b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)
c)x^2-2x-5x+10=x(x-2)-5(x-2)=(x-5)(x-2)
d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(x-2)
a/ (5x^2 -10xy+5y^2)-20z^2=5[(x-y)^2-(2z)^2]=5(x-y-2x)(x-y+2z)
b/16x-5x^2-3=-[5x^2-16x+3]=-[(5x^2-x)-(15x+3)]=-[x(5x-1)-3(5x-1)]=(3-x)(5x-1)
c/x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)
2a/ 5x(X^2-9)=0=>x=0 hoặc x^2=9=>x=0 hoặc x=+-3
b/x^2-7x+10=0=>(x^2-2x)-(5x-10)=0=>x(x-2)-5(x-2)=0=>x-2=0 hoặc x-5 =0 => tự tính nhé!
Answer:
Bài 1:
\(5x² - 10xy + 5y² - 20z²\)
\(= 5( x² - 2xy + y² - 4z²)\)
\(= 5 [(x² - 2xy + y²) - (2z)²]\)
\(= 5 [(x - y)² - (2z)²]\)
\(= 5 (x - y - 2z) ( x - y + 2z)\)
\(16x - 5x² - 3 \)
\(= -( 5x² - 16x + 3)\)
\(= -( 5x² - 15x - 1x + 3)\)
\(= - [ (5x² -x) - (15x -3)]\)
\(= - [ x(5x -1) -3(5x -1)]\)
\(= - (5x-1)(x-3)\)
\(x² + 4x + 3\)
\(= x² + x + 3x + 3\)
\(= (x² + x) + (3x + 3)\)
\(= x( x + 1) +3 (x+1)\)
\(= (x+1) (x+3)\)
Bài 2:
\(5x\left(x^2-9\right)=0\)
\(\Rightarrow5x\left(x-3\right)\left(x+3\right)=0\)
Trường hợp 1: \(5x=0\Leftrightarrow x=0\)
Trường hợp 2: \(x-3=0\Leftrightarrow x=3\)
Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)
\(x^2-7x+10=0\)
\(\Rightarrow x^2-5x-2x+10=0\)
\(\Rightarrow x\left(x-5\right)-2\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)
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)
x2.(1 - x2) - 4 + 4x2
= x2.(1 - x2) - 4.(1 - x2)
= (1 - x2).(x2 - 4)
= (1 - x)(1 + x)(x - 2)(x + 2)
dung roi