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a) 6x2 - 12x
= 6x(x - 2)
b) x2 + 2x + 1 - y2
= (x2 + 2x + 1) - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
c) x + y + z + x2 + xy + xz
= (x + x2) + (y + xy) + (z + xz)
= x(1 + x) + y(1 + x) + z(1 + x)
= (x + y + z)(x + 1)
d) xy + xz + y2 + yz
= (xy + xz) + (y2 + yz)
= x(y + z) + y(y + z)
= (x + y)(x + z)
e) x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x2 + 1)(x + 1)
f) xy + y - 2x - 2
= (xy + y) - (2x + 2)
= y(x + 1) - 2(x + 1)
= (y - 2)(x + 1)
g) x3 + 3x - 3x2 - 9
= (x3 - 3x2) + (3x - 9)
= x2(x - 3) + 3(x - 3)
= (x2 + 3)(x - 3)
h) x2 - y2 - 2x - 2y
= (x2 - y2) - (2x + 2y)
= (x + y)(x - y) - 2(x + y)
= (x + y)(x - y - 2)
i) 7x2 - 7xy - 5x = 5y
mk thấy con này sai sai ý
1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)
\(=4x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(4x+1\right)\)
2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)
\(=4x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(4x-3\right)\)
3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)
\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)
4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=xy\left(x+y\right)-y^2z-yz^2+x^2z-xz^2\)
\(=xy\left(x+y\right)+\left(x^2z-y^2z\right)-\left(yz^2+xz^2\right)\)
\(=xy\left(x+y\right)+z\left(x^2-y^2\right)-z^2.\left(x+y\right)\)
\(=xy\left(x+y\right)+z\left(x-y\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=xy\left(x+y\right)+\left(zx-zy\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)
1) 4x2 - 7x - 2 = 4x2 - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )
2) 4x2 + 5x - 6 = 4x2 - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )
3) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
4) xy( x + y ) - yz( y + z ) + xz( x - z )
= x2y + xy2 - y2z - yz2 + xz( x - z )
= ( x2y - yz2 ) + ( xy2 - y2z ) + xz( x - z )
= y( x2 - z2 ) + y2( x - z ) + xz( x - z )
= y( x - z )( x + z ) + y2( x - z ) + xz( x - z )
= ( x - z )[ y( x + z ) + y2 + xz ]
= ( x - z )( xy + yz + y2 + xz )
= ( x - z )[ ( xy + y2 ) + ( xz + yz ) ]
= ( x - z )[ y( x + y ) + z( x + y ) ]
= ( x - z )( x + y )( y + z )
5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )
1)2xy+3z+6y+xz
= x(2y + z) + 3(z + 2y)
= (x + 3)(2y + z)
2)x^4-9x^3+x^2-9x
= x^2(x^2 + 1) - 9x(x^2 + 1)
= (x^2 + 1)(x^2 - 9x)
= x(x - 9)(x^2 + 1)
3)x^2-xy+x-y
= x(x - y) + (x - y)
= (x + 1)(x - y)
4)xz+yz-5(x+y)
= z(x + y) - 5(x + y)
= (z - 5)(x + y)
5)3x^2-3xy-5x+5y
= 3x(x - y) - 5(x - y)
= (3x - 5)(x - y)
6)x^2+4x-y^2+4y
= (x - y)(x + y) + 4(x + y)
= (x - y + 4)(x + y)
\(b,9x^2+90x+225-\left(x-y\right)^2\)
\(=\left(3x+15\right)^2-\left(x-y\right)^2\)
\(=\left(3x+15-x+y\right)\left(3x+15+x-y\right)\)
\(=\left(2x+y+15\right)\left(4x-y+15\right)\)
1: \(=a\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(a-4\right)\)
2: \(=x\left(x+b\right)+a\left(x+b\right)=\left(x+b\right)\left(x+q\right)\)
3: \(=a\left(x+1\right)-b\left(x+1\right)+c\left(x+1\right)\)
\(=\left(x+1\right)\left(a-b+c\right)\)
6: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1-x+3\right)^2=\left(x+4\right)^2\)
b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)
c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)
d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y+z\right)\left(x+y\right)\)
f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
1,
a, (2x + 1- x + 3)2 = (x+4)2
b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)
c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)
d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)
e,\(\left(x+y\right)^2-z\left(x+y\right)\)=\(\left(x+y\right)\left(x+y-z\right)\)
f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)
\(=\left(5+2x+y\right)\left(5-2x-y\right)\)
Chúc các bn hc tốt
B3) a) x(x-5)-4(x-5)=0
<=> (x-4)(x-5)=0
TH1 :x-4=0
<=.x=4
TH2 : x-5=0
<=>x=5
b) x(x-6)-7x-42=0
<=>x(x+6)-7(x+6)=0
<=>(x-7)(x+6)=0
th1;x-7=0
<=>x=7
th2; x+6=0
<=>x=-6
c)x^3-5x^2+x-5=0
<=> x(x^2+1)-5(x^2+1)=0
<=> (x-5)(x^2+1)=0
th1:x-5=0
<=>x=5
TH2 : x^2+1=0
<=> x^2=-1 ( vo li )
=> th2 ko tồn tại
nho thick nha
Bài 3
a, x(x-5)-4(x-5)=0
(x-4)(x-5)=0
=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
b,x(x+6)-7(x+6)=0
(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
c,x^2(x-5)+(x-5)=0
(x^2+1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
Thời gian có hạn copy cái này hộ mình vào google xem nha: :
Link : https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi
Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....
Có 500 giải nhanh nha đã có 200 người nhận rồi. Mình là phụ trách
OK N