Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,4x^4-8x^3+4x^2\)
\(=4x^2\cdot\left(x^2-2x+1\right)\)
\(=4x^2\cdot\left(x-1\right)^2\)
\(b,x^2-y^2+5\cdot\left(y-x\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x+y-5\right)\)
\(c,3x^2-6xy+3y^2-12z^2\)
\(=3\cdot\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\cdot\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)
\(=3\cdot\left(x-y-2x\right)\cdot\left(x-y+2x\right)\)
a) \(ax+ay-3x-3y=a\left(x+y\right)-3\left(x+y\right)=\left(a-3\right)\left(x+y\right)\)
b) \(x^3-3x^2+3x-9=x^2\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x^2+3\right)\)
c) xem lại đề
d) \(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3-x-y\right)\left(3+x+y\right)\)
a) ( x2 - 25 )2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) ]2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) - ( x - 5 ) ][ ( x - 5 )( x + 5 ) + ( x - 5 ) ]
= ( x - 5 )( x + 5 - 1 )( x - 5 )( x + 5 + 1 )
= ( x - 5 )2( x + 4 )( x + 6 )
b) ( 4x2 - 25 )2 - 9( 2x - 5 )2
= ( 4x2 - 25 )2 - 32( 2x - 5 )2
= ( 4x2 - 25 )2 - ( 6x - 15 )2
= [ ( 4x2 - 25 ) - ( 6x - 15 ) ][ ( 4x2 - 25 ) + ( 6x - 15 ) ]
= ( 4x2 - 25 - 6x + 15 )( 4x2 - 25 + 6x - 15 )
= ( 4x2 - 6x - 10 )( 4x2 + 6x - 40 )
= ( 4x2 + 4x - 10x - 10 )( 4x2 + 16x - 10x - 40 )
= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]
= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )
= ( 4x - 10 )2( x + 1 )( x + 4 )
c) 4( 2x - 3 )2 - 9( 4x2 - 9 )2
= 22( 2x - 3 )2 - 32( 4x2 - 9 )2
= ( 4x - 6 )2 - ( 12x2 - 27 )2
= [ ( 4x - 6 ) - ( 12x2 - 27 ) ][ ( 4x - 6 ) + ( 12x2 - 27 ) ]
= ( 4x - 6 - 12x2 + 27 )( 4x - 6 + 12x2 - 27 )
= ( -12x2 + 4x + 21 )( 12x2 + 4x - 33 )
= ( -12x2 + 18x - 14x + 21 )( 12x2 - 18x + 22x - 33 )
= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]
= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )
= ( x - 3/2 )2( -12x - 14 )( 12x + 22 )
d) a6 - a4 + 2a3 + 2a2
= a2( a4 - a2 + 2a + 2 )
= a2( a4 - 2a3 + 2a3 + 2a2 - 4a2 + a2 + 4a - 2a + 2 )
= a2[ ( a4 - 2a3 + 2a2 ) + ( 2a3 - 4a2 + 4a ) + ( a2 - 2a + 2 ) ]
= a2[ a2( a2 - 2a + 2 ) + 2a( a2 - 2a + 2 ) + 1( a2 - 2a + 2 ) ]
= a2( a2 + 2a + 1 )( a2 - 2a + 2 )
= a2( a + 1 )2( a2 - 2a + 2 )
e) ( 3x2 + 3x + 2 )2 - ( 3x2 + 3x - 2 )2
= [ ( 3x2 + 3x + 2 ) - ( 3x2 + 3x - 2 ) ][ ( 3x2 + 3x + 2 ) + ( 3x2 + 3x - 2 ) ]
= ( 3x2 + 3x + 2 - 3x2 - 3x + 2 )( 3x2 + 3x + 2 + 3x2 + 3x - 2 )
= 4( 6x2 + 6x )
= 4.6x( x + 1 )
= 24( x + 1 )
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
a) \(C=ax-ay-bx+by+\left(y-x\right)^2\)
\(=a\left(x-y\right)-b\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(a-b+x-y\right)\)
b) \(M=\left(5x-10\right)\left(x^2-1\right)\left(3x-6\right)\left(x^2-2x+1\right)\)
\(=5\left(x-2\right)\left(x-1\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)^2\)
\(=\left(x-2\right)\left(x-1\right)\left[5\left(x+1\right)-3\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(5x+5-3x+3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(2x+8\right)\)
\(=2\left(x-2\right)\left(x-1\right)\left(x+4\right)\)
a)Bt = (x2-a2)-(2x-2a)
=....
b)Bấm máy tìm nghiệm đi rồi phân tích
c);d);e);f)Nhóm số đầu vs số thứ 2, số thứ 3 vs số thứ 4
1.a)\(x^2-ax+bx-ab=x\left(x-a\right)+b\left(x-a\right)=\left(x+b\right)\left(x-a\right)\)
b)\(x^2+ay-y^2-ax=\left(x-y\right)\left(x+y\right)-a\left(x-y\right)=\left(x+y-a\right)\left(x-y\right)\)
c)\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2.a)\(2x^2-12x=-18=>2x^2-12x+18=0=>x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)b)\(\left(4x^2-4x+1\right)-x^2=0=>3x^2-3x-x+1=3x\left(x-1\right)-\left(x-1\right)=\left(3x-1\right)\left(x-1\right)=0\)
\(=>\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}=>\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)
a) 2x2 - 12x = -18
<=> 2x2 - 12x + 18 = 0
<=> 2(x2 - 6x + 9) = 0
<=> 2(x2 - 2.x.3 + 9) = 0
<=> 2(x - 3)2 = 0
<=> x - 3 = 0
<=> x = 0 + 3
<=> x = 3
b) (4x2 - 4x + 1) - x2 = 0
<=> 4x2 - 4x + 1 - x2 = 0
<=> 3x2 - 4x + 1 = 0
<=> 3x2 - x - 3x + 1 = 0
<=> x(3x - 1) - (3x - 1) = 0
<=> \(\orbr{\begin{cases}\left(3x-1\right)=0\\\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)