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Bài 1 :
a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)
b) \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)
c) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
d) \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)
BÀi 2 :
a) \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)
\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)
b) \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)
c) \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)
\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)
\(=\left(b+c-a\right)\left(d-c^2\right)\)
BÀi 3 :
a) \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)
b) \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)
c) \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)
d) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
Bài 1
1) 4x - x2 - 4 = 0
⇔ -( x2 - 4x + 4 ) = 0
⇔ -( x - 2 )2 = 0
⇔ x - 2 = 0
⇔ x = 2
2) 4( x - 1 )2 - ( 5 - 2x )2 = 0
⇔ 22( x - 1 )2 - ( 5 - 2x )2 = 0
⇔ ( 2x - 2 )2 - ( 5 - 2x ) = 0
⇔ ( 2x - 2 - 5 + 2x )( 2x - 2 + 5 - 2x ) = 0
⇔ ( 4x - 7 ).3 = 0
⇔ 4x - 7 = 0
⇔ x = 7/4
3) 9( x - 2 )2 - 4( 3 - x )2 = 0
⇔ 32( x - 2 )2 - 22( x - 3 )2 = 0
⇔ ( 3x - 6 )2 - ( 2x - 6 )2 = 0
⇔ ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 ) = 0
⇔ x( 5x - 12 ) = 0
⇔ x = 0 hoặc 5x - 12 = 0
⇔ x = 0 hoặc x = 12/5
4) x2 - 6x + 5 = 0
⇔ x2 - 5x - x + 5 = 0
⇔ x( x - 5 ) - ( x - 5 ) = 0
⇔ ( x - 5 )( x - 1 ) = 0
⇔ x - 5 = 0 hoặc x - 1 = 0
⇔ x = 5 hoặc x = 1
Bài 2.
1) x2 - z2 + y2 - 2xy
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z )( x - y + z )
2) a3 - ay - a2x + xy
= ( a3 - a2x ) - ( ay - xy )
= a2( a - x ) - y( a - x )
= ( a - x )( a2 - y )
3) 2xy + 3z + 6y + xz
= ( 2xy + 6y ) + ( xz + 3z )
= 2y( x + 3 ) + z( x + 3 )
= ( x + 3 )( 2y + z )
4) x2 + 2xz + 2xy + 4yz
= ( x2 + 2xy ) + ( 2xz + 4yz )
= x( x + 2y ) + 2z( x + 2y )
= ( x + 2y )( x + 2z )
5) ( x + y + z )3 - x3 - y3 - z3
= x3 + y3 + z3 + 3( x + y )( y + z )( x + z ) - x3 - y3 - z3
= 3( x + y )( y + z )( x + z )
a) 4x2-8x=0
(2x)2-2.2.2x+4-4=0
(2x-2)2 =4
2x-2=2
2x =4
x=2
Nhớ k cho mk nha
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
Bài 1 :
a) xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
b) \(x^3-x+3x^2y+3xy^2+y^3-x-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Đã có kết quả
Bài 1,chữa phần a
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
=[xy(x+y)+xyz]+[yz(y+z)+xyz]+xz(x+z)
=xy(x+y+z)+yz(x+y+z)+xz(x+z)
=y(x+y+z)(x+z)+xz(x+z)
=(x+z)(xy+y2+yz+xz)
=(x+z)(x+y)(y+z)
Chữa phần b
x3-x+3x2y+3xy2+y3-y
=(x+y)(x+y-1)(x+y+1)
Bài2
a3+b3+c3=(a+b)3-3ab(a+b)+c3=-c3-3ab(-c)+c3=3abc
Ai làm đúng như này ớ sẽ k
BÀI 1.
a. 2.( x+5 ) - x2 -5x = 2. (x+5) - x .(x +5 )
=( x+5 ). (2 - x)
b. y2 - 6y +9 +z2 =( y2 -6y +9 )+ z2
=(y - 3)2 +z2
c. a3 - a2x- ay +xy =( a3 - a2x) - (ay - xy )
=a2 (a-x) - y (a -x)
=(a - x) . (a2 - y)
bài 2
a. x2 - 6x =0
x( x -6 ) =0
Suy ra : x= 0 hoặc x- 6 =0
1) x =0
2) x -6 =0 suy ra x=6
vậy x =0 ; x= 6
b. x3 -2x2 +x =0
x . ( x2 - 2x +1 ) =0
x . ( x -1 )2 =0
suy ra : x = 0 hoặc (x - 1)2 =0
1) x = 0
2) (x - 1)2 = 0 suy ra x -1 = 0
suy ra : x= 1
vậy x = 0 ; x = 1
Tick cho mk nhé!!!!!!!