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a)5x.(x-2y)+2.(2y-x)2 = 5x.(x-2y)+2.(x-2y)2 = (x-2y)[ 5x+2(x-2y)] = (x-2y)( 5x + 2x - 4y) = (x-2y)(7x-4y)
b) tương tự như trước, do là (A-B)^2 = a^2 - 2ab + b^2 = b^2 - 2ab + a^2 = ( b-a)^2
c)100x2-(x2+25)2= (10x)^2 -(x2+25)2= (10x + x^2 + 25)( 10x-x^2-25) = (x^2+10x+25)[-(x^2-10x+25)] =-1(x+5)^2 (x-5)^2
d)x2-xz-9y2+3y2= ( từ để suy nghĩ đã -_-)
e)x3-x2.5x+125 =
a) x3 - 1 + 5x2 - 5 + 3x - 3
= x3 + 5x2 + 3x - 9
= x3 + 6x2 - x2 + 9x - 6x - 9
= ( x3 + 6x2 + 9x ) - ( x2 + 6x + 9 )
= x( x2 + 6x + 9 ) - ( x2 + 6x + 9 )
= ( x2 + 6x + 9 )( x - 1 )
= ( x + 3 )2( x - 1 )
b) a5 + a4 + a3 + a2 + a + 1
= ( a5 + a4 + a3 ) + ( a2 + a + 1 )
= a3( a2 + a + 1 ) + 1( a2 + a + 1 )
= ( a2 + a + 1 )( a3 + 1 )
= ( a2 + a + 1 )( a + 1 )( a2 - a + 1 )
c) x3 - 3x2 + 3x - 1 - y3
= ( x3 - 3x2 + 3x - 1 ) - y3
= ( x - 1 )3 - y3
= ( x - 1 - y )[ ( x - 1 )2 + ( x - 1 )y + y2 ]
= ( x - 1 - y )( x2 - 2x + 1 + xy - y + y2 )
d) 5x3 - 3x2y - 45xy2 + 27y3
= ( 5x3 - 45xy2 ) - ( 3x2y - 27y3 )
= 5x( x2 - 9y2 ) - 3y( x2 - 9y2 )
= ( 5x - 3y )( x2 - 9y2 )
= ( 5x - 3y )[ x2 - ( 3y )2 ]
= ( 5x - 3y )( x - 3y )( x + 3y )
a)x4-4(x2+5)-25=x4-4x2-45=(x4-9x2)+(5x2-45)=x2(x2-9)+5(x2-9)=(x2-9)(x2+5)=(x-3)(x+3)(x2+5)
b)a2-b2-2a+1=(a2-2a+1)-b2=(a-1)2-b2=(a-b-1)(a+b-1)
c)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
d)x2+4x-y2+4=(x2+4x+4)-y2=(x+2)2-y2=(x-y+2)(x+y+2)
a)\(6x^2-9xy\)
\(=3x\left(2x-3y\right)\)
b)\(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
c)\(x^4-8x^2-9\)
\(=x^4+x^2-9x^2-9\)
\(=x^2\left(x^2+1\right)-9\left(x^2+1\right)\)
\(=\left(x^2-9\right)\left(x^2+1\right)\)
\(=\left(x+3\right)\left(x-3\right)\left(x^2+1\right)\)
d)\(x^4-4\left(x^2+5\right)-25\)
\(=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-5-4\right)\)
\(=\left(x^2+5\right)\left(x^2-9\right)\)
\(=\left(x^2+5\right)\left(x-3\right)\left(x+3\right)\)
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
\(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(x^2+10x+25\right)\)
\(=-\left(x-5\right)^2\left(x+5\right)^2\)
\(b,x-y+5=a\text{ thì biểu thức bằng:}a^2-2a+4>0\text{ nên k phân tích đc}\)
\(d,x^3+27y^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
a) 100x2 - ( x2 + 25 )2
= ( 10x )2 - ( x2 + 25 )2
= [ 10x - ( x2 + 25 ) ][ 10x + ( x2 + 25 ) ]
= ( -x2 + 10x - 25 )( x2 + 10x + 25 )
= -( x2 - 10x + 25 )( x2 + 10x + 25 )
= -( x - 5 )2( x + 5 )2
b) ( x - y + 5 )2 + 4 - 4( x - y + 5 ) ( 4 may ra còn phân tích được :)) )
= ( x - y + 5 )2 - 2( x - y + 5 ).2 + 22
= ( x - y + 5 - 2 )2
= ( x - y + 3 )2
c) a2 - 25( b - c ) ( không phân tích được :)) )
d) x3 + 27y3 = x3 + ( 3y )3 = ( x + 3y )( x2 - 3xy + 9y2 )