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27 tháng 7 2021
1, \(2\left(x+y\right)-5a\left(x+y\right)=\left(x+y\right)\left(2-5a\right)\)
2, \(a^2\left(x-5\right)-3\left(x-5\right)=\left(a^2-3\right)\left(x-5\right)\)
3, \(4x\left(a-b\right)+6xy\left(b-a\right)=\left(4x-6xy\right)\left(a-b\right)=2x\left(2-3y\right)\left(a-b\right)\)
4, \(y\left(a-b\right)-x\left(b-a\right)=\left(x+y\right)\left(a-b\right)\)
5, \(6x\left(x-y\right)+8y\left(y-x\right)=\left(x-y\right)\left(6x-8y\right)=2\left(3x-4y\right)\left(x-y\right)\)
6, \(4\left(x-3\right)^2-2x\left(x-3\right)=\left(x-3\right)\left[4\left(x-3\right)-2x\right]=2\left(x-3\right)\left(x-6\right)\)
21 tháng 3 2020
1)\(4\left(a^4-1\right)x=5\left(a-1\right)\)
<=>x=\(\frac{5\left(a-1\right)}{a^4-1}\)
<=>x=\(\frac{5\left(a-1\right)}{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}=\frac{5}{\left(a+1\right)\left(a^2+1\right)}\)
Tương tự ta tính được y=\(\frac{4a^6+4}{5a^4-5a^2+5}\)
Suy ra x.y=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\cdot\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)
=\(\frac{5}{a+1}\)
Tương tự với x:y
21 tháng 3 2020
\(A=\frac{4.6}{4.2}:\left(\frac{8.10}{6.8}.\frac{12.14}{10.12}.\frac{16.18}{14.16}...\frac{54.56}{54.53}\right)=\frac{6}{2}:\frac{56}{6}=\)
D
26 tháng 8 2019
1, 5a2xy-10a3x-15ay = 5a( axy - 2a\(^2\)x - 3y )
2, mxy-m2x+my = m( xy - mx + y )
3, 2mx-4m2xy+6mx = 2mx( 1 - 2my + 3 ) = 2mx( 4 - 2my )
4, a2b-2ab2+ab = ab( a - 2b + 1 )
5, 5a2b-2ab2+ab = ab( 5a - 2b +1 )
6, -3x2y3-6x3y2-x2y2 = -3x\(^2\)y\(^2\) ( y + 2x + 1 )
7, 5x2y4-10x4y2+5x2y2 = 5x\(^{^2y^2}\)( y\(^2\) - 2x\(^2\) + 1 )
8, -2x3y4-4x4y3+2x3y3 = 2\(x^3y^3\) ( -y - 2x + 1 )
9, 4x3y2-8x3y+16xy2-24 = 4( x\(^3\)y\(^2\) - 2x\(^3\)y + 4 xy\(^2\) - 6 )
10, 12x3y-6xy+3x = 3x( 4x\(^2\)y - 2y + 1 )
11, 2(x-y)-a(x-y) = ( 2 - a ) ( x - y )
12, a(x-y)+b(x-y)= ( a + b ) ( x - y )
13, m(x+y)-n(x+y) = ( m - n ) ( x + y )
14, 2a(x+y)-4(x+y) = ( 2a - 4 )( x + y ) = 2( a - 2 ) ( x + y )
15, 3a(x+y)-6ab(x+y) = ( 3a - 6ab )( x + y ) = 3a( 1 - 2b ) ( x + y )
16, 5a2(x-y)+10a(x-y) = ( 5a\(^2\)+10a )( x - y ) = 5a( a + 2 ) ( x - y )
17, -2ab(x-y)-4a(x-y) = ( -2ab - 4a )( x - y ) = -2a( b + 2 )( x - y )
18, 3a(x-y)+2(x-y) = ( 3a + 2 ) ( x - y )
19, m(a-b)-m2(a-b) = ( m - m\(^2\) ) ( a - b ) = m( 1 - m ) ( a - b )
20, mx(a+b)-m(a+b) = ( mx - m ) ( a + b ) = m( x - 1 )( a + b )
21, x(a-b)-y(b-a) = x( a - b ) + y( a - b ) = ( x + y ) ( a - b )
22, ab(x-5)-a2(5-x) = ab( x - 5 ) + a\(^2\)( x - 5 ) = ( ab + a\(^2\) ) ( x - 5 ) = a( b + a )( x - 5 )
23, 2a2(x-y)-4a(y-x)= 2a\(^2\)( x - y ) + 4a( x - y )=( 2a\(^2\) + 4a ) ( x - y )= 2a( a + 2 )( x - y )
X
26 tháng 8 2019
Đăng ít thôi =))
a. \(5a^2xy-10a^3x-15ay=5a\left(axy-2a^2x-3y\right)\)
b. \(mxy-m^2x+my=m\left(xy-mx+y\right)\)
c. \(2mx-4m^2xy+6mx=2mx\left(1-2my+3\right)=2mx\left(-2my+4\right)\)
d. \(a^2b-2ab^2+ab=ab\left(a-2b+1\right)\)
e. \(5a^2b-2ab^2+ab=ab\left(5a-2b+1\right)\)
g.
MN
8 tháng 8 2020
a) Sửa đề :
\(x^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(x^4=\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2+3ab^3+b^4\right)\)
\(x^4=a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(x^4=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(x^4=\left(a+b\right)\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)
\(x^4=\left(a+b\right)\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)
\(x^4=\left(a+b\right)^2\left(a+2ab+b^2\right)\)
\(x^4=\left(a+b\right)^4\)
b) Sửa đề:
\(x^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)
\(x^5=\left(a^5+4a^4b+6a^3b^2+4a^2b^3+ab^4\right)+\left(a^4b+4a^3b^2+6a^2b+4ab^4+b^5\right)\)
\(x^5=a\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)+b\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)
\(x^5=\left(a+b\right)\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)
\(x^5=\left(a+b\right)\left[\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2++3ab^3+b^4\right)\right]\)
\(x^5=\left(a+b\right)\left[a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\right]\)
\(x^5=\left(a+b\right)^2\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(x^5=\left(a+b\right)^2\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)
\(x^5=\left(a+b\right)^2\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)
\(x^5=\left(a+b\right)^3\left(a^2+2ab+b^2\right)\)
\(x^5=\left(a+b\right)^5\)
Bạn có thể tự tóm tắt lại
28 tháng 10 2017
Bài 2 :
f(x) có bậc 3 chia cho đa thức \(x^2-x-2\) có bậc 2 sẽ được thương có bậc 1
Gọi thương của phép chia f(x) cho \(x^2-x-2\) là \(cx+d\)
\(\left(cx+d\right)\left(x^2-x-2\right)=f\left(x\right)\)
hay \(cx^3-cx^2-2cx+dx^2-dx-2d=x^3+ax+b\)
\(\Rightarrow cx^3+\left(d-c\right)x^2-\left(2c+d\right)x-2d=x^3+ax+b\)
\(\Rightarrow\left\{{}\begin{matrix}cx^3=x^3\\\left(d-c\right)x^2=0\\-\left(2c+d\right)x=ax\\-2d=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=1\\d-1=0\\a=-2.1-d\\-2d=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=1\\d=1\\a=-3\\b=-2\end{matrix}\right.\)
TM
7 tháng 8 2017
dài quá, làm từ từ nhé
1, \(\left(a-b\right)^2\left(2a-3b\right)-\left(b-a\right)^2\left(3a-5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(2a-3b-3a+5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(-a+2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=-\left(a-b\right)^2\left(a-2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-2b\right)\left[\left(a+b\right)^2-\left(a-b\right)^2\right]\)
\(=\left(a-2b\right)\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\left(a-2b\right)\)
2, \(x^4-4\left(x^2+5\right)-25=\left(x^2-25\right)-4\left(x^2+5\right)=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2-9\right)\left(x^2+5\right)=\left(x-3\right)\left(x+3\right)\left(x^2+5\right)\)
TM
7 tháng 8 2017
3,\(\left(2-x\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)=\left(x-2\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)\)
\(=\left(x-2\right)\left(x-2+x+3\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2\right)\left(2x+1\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2-2x+1\right)\left(2x+1\right)\)
\(=\left(-x-1\right)\left(2x+1\right)\)
4, câu này đề thiếu
5,\(16\left(xy+6\right)^2-\left(4x^2+y^2-25\right)^2=\left(4xy+24\right)^2-\left(4x^2+y^2-25\right)^2\)
\(=\left(4xy+24-4x^2-y^2+25\right)\left(4xy+24+4x^2+y^2-25\right)\)
\(=\left[49-\left(4x^2-4xy+y^2\right)\right]\left[\left(4x^2+4xy+y^2\right)-1\right]\)
\(=\left[49-\left(2x-y\right)^2\right]\left[\left(2x+y\right)^2-1\right]\)
\(=\left(7-2x+y\right)\left(7+2x-y\right)\left(2x+y-1\right)\left(2x+y+1\right)\)
ai trả lời đc mình cho
ĐKXĐ bạn tự xét nha
a) \(4\left(a^4-1\right)x=5\left(a-1\right)\)
\(\Leftrightarrow x=\frac{5\left(a-1\right)}{4\left(a^4-1\right)}\)
\(\Leftrightarrow x=\frac{5\left(a-1\right)}{4\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}\)
\(\Leftrightarrow x=\frac{5}{4\left(a+1\right)\left(a^2+1\right)}\)
b) \(\left(5a^4-5a^2+5\right)y=4a^6+4\)
\(\Leftrightarrow y=\frac{4a^6+4}{5a^4-5a^2+5}\)
\(\Leftrightarrow y=\frac{4\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)
\(\Leftrightarrow y=\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)
\(\Leftrightarrow y=\frac{4\left(a^2+1\right)}{5}\)