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21 tháng 7 2019
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)\(\left(4x^2-4x+1\right)-x^2=0=>3x^2-3x-x+1=3x\left(x-1\right)-\left(x-1\right)=\left(3x-1\right)\left(x-1\right)=0\)
\(=>\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}=>\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)
21 tháng 7 2019
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
<=> \(\orbr{\begin{cases}\left(3x-1\right)=0\\\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
TN
19 tháng 8 2017
\(a,7x^2-7xy-4x+4y\)
\(=7x\left(x-y\right)-4\left(x-y\right)\)
\(=\left(7x-4\right)\left(x-y\right)\)
\(b,2x-2y+ax-ay\)
\(=2\left(x-y\right)+a\left(x-y\right)\)
\(=\left(a+2\right)\left(x-y\right)\)
\(c,x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
\(d,ax+ay-2x-2y\)
\(=a\left(x+y\right)-2\left(x+y\right)\)
\(=\left(a-2\right)\left(x+y\right)\)
\(e,x\left(a+b\right)-a-b=x\left(a+b\right)-\left(a+b\right)\)
\(=\left(x-1\right)\left(a+b\right)\)
NC
19 tháng 8 2020
Bài làm:
a) \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(\left(x-y\right)\left(x-y-z\right)\)
A
19 tháng 8 2020
a/ \(x^2-2xy+y^2-zx+yz.\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c/ \(x^2-y^2-2x-2y.\)
\(=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
1 tháng 7 2017
1/Tự chép lại đb nha :v
=a2 - 9b2+2ab+3a2-8b2-12ab+6ab-3b2-2a2+ab
= 2a2-3ab-20b2
= (2a2+5ab) - (8ab+20b2)
= a(2a+5b) - 4b(2a+5b)
=(2a+5b)(a-4b)
câu 2 tương tự nhé :)
TK
27 tháng 10 2019
a) x3+y3+2x2-2xy+2y2=(x3+y3)+(2x2-2xy+2y2)=(x+y)(x2-xy+y2)+2(x2-xy+y2)=(x+y+2)(x2-xy+y2)
b) tương tự câu a
TK
27 tháng 10 2019
c) (a2+b2-5)2-4(ab+2)2=(a2+b2-5)2-22(ab+2)2
= (a2+b2-5)2-(2ab+4)2=(a2-2ab+b2-5-4)(a2+2ab+b2-5+4)
=( (a-b)2 -9 )( (a+b)2 -1 ) = (a-b-3)(a-b+3)(a+b-1)(a+b+1)
12 tháng 10 2018
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
12 tháng 10 2018
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
TA
25 tháng 7 2017
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
8 tháng 9 2020
A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )
= 4x( ac + bc + a + b )
= 4x[ c( a + b ) + ( a + b ) ]
= 4x( a + b )( c + 1 )
B = ax - bx + cx - 3a + 3b - 3c
= x( a - b + c ) - 3( a - b + c )
= ( a - b + c )( x - 3 )
C = 2ax - bx + 3cx - 2a + b - 3c
= x( 2a - b + 3c ) - ( 2a - b + 3c )
= ( 2a - b + 3c )( x - 1 )
D = ax - bx - 2cx - 2a + 2b + 4c
= x( a - b - 2c ) - 2( a - b - 2c )
= ( a - b - 2c )( x - 2 )
E = 3ax2 + 3bx2 + ax + bx + 5a + 5b
= 3x2( a + b ) + x( a + b ) + 5( a + b )
= ( a + b )( 3x2 + x + 5 )
F = ax2 - bx2 - 2ax + 2bx - 3a + 3b
= x2( a - b ) - 2x( a - b ) - 3( a - b )
= ( a - b )( x2 - 2x - 3 )
= ( a - b )( x2 + x - 3x - 3 )
= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]
= ( a - b )( x + 1 )( x - 3 )