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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 )
1) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
2)\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
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
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
Bài 1 :
Câu a : \(a^3+a^2b-a^2c-abc\)
\(=a\left(a^2+ab-ac-bc\right)\)
\(=a\left[a\left(a+b\right)-c\left(a+b\right)\right]\)
\(=a\left(a+b\right)\left(a-c\right)\)
Câu b : \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
Câu c : \(4-x^2-2xy-y^2\)
\(=4-\left(x^2+2xy+y^2\right)\)
\(=2^2-\left(x+y\right)^2\)
\(=\left(2-x-y\right)\left(2+x+y\right)\)
Câu d : \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
Bài 2 :
Câu a : \(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=-\dfrac{1}{2}\) hoặc \(x=3\)
Câu b : \(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=2\)
Câu c : \(x\left(x-1\right)+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(x=-2\) hoặc \(x=1\)
Câu d : \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=-\dfrac{3}{2}\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
bạn giải lại giúp mình bài 2 được ko ạ