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BR
4
BR
1
VB
30 tháng 7 2017
\(2x^3+2ax-4a-4x\\ =2\left(x^3+ax-2a-2x\right)\)
câu sau thì chịu
B
2
19 tháng 7 2018
\(mx^2-4mx+4m-nx^2+4nx-4n\)
\(=\left(mx^2-4mx+4m\right)-\left(nx^2-4nx+4n\right)\)
\(=m\left(x^2-4x+4\right)-n\left(x^2-4x+4\right)\)
\(=\left(m-n\right)\left(x^2-4x+4\right)\)
\(=\left(m-n\right)\left(x-2\right)^2\)
AK
19 tháng 7 2018
\(mx^2-4mx+4m-nx^2+4nx-4n\)
\(=x^2\left(m-n\right)+4x\left(n-m\right)+4\left(m-n\right)\)
\(=x^2\left(m-n\right)-4x\left(m-n\right)+4\left(m-n\right)\)
\(=\left(x^2-4x+4\right)\left(m-n\right)\)
\(=\left(x-2\right)^2\left(m-n\right)\)
2T
4 tháng 9 2019
\(1.mx+my+5x+5y\)
\(=m\left(x+y\right)+5\left(x+y\right)\)
\(=\left(m+5\right)\left(x+y\right)\)
\(2.\left(a+b\right)\left(a-b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a-b-a^2+ab-b^2\right)\)
G
3
18 tháng 12 2019
a) \(mx^2-4mx+4m-nx^2+4nx-4n\)
\(=\left(mx^2-nx^2\right)-\left(4mx-4nx\right)+\left(4m-4n\right)\)
\(=x^2\left(m-n\right)-4x\left(m-n\right)+4\left(m-n\right)\)
\(=\left(m-n\right)\left(x^2-4x+4\right)\)
\(=\left(m-n\right)\left(x-2\right)^2\)
b) \(3x^2+48+24x-12y^2\)
\(=3\left(x^2+8x+16-4y^2\right)\)
\(=3\left[\left(x+4\right)^2-\left(2y\right)^2\right]\)
\(=3\left(x+4-2y\right)\left(x+4+2y\right)\)
HN
18 tháng 12 2019
a) \(mx^2-4mx+4m-nx^2+4nx-4n\)
\(=\left(mx^2-nx^2\right)-\left(4mx+4nx\right)+\left(4m-4n\right)\)
\(=x^2\left(m-n\right)-4x\left(m-n\right)+4\left(m-n\right)\)
\(=\left(m-n\right).\left(x^2-4x+4\right)\)
\(=\left(m-n\right).\left(x-2\right)^2\)
b) \(3x^2+48+24x-12y^2\)
\(=3\left(x^2+16+8x-4y^2\right)\)
\(=3\left[\left(x^2+8x+16\right)-\left(2y\right)^2\right]\)
\(=3\left[\left(x+4\right)^2-\left(2y\right)^2\right]\)
\(=3\left(x+4-2y\right).\left(x+4+2y\right)\)
17 tháng 11 2017
Bài 1: Phân tích đa thức thành nhân tử:
a) x(y+z) + 3(y+z)
\(=\left(y+z\right)\left(x+3\right)\)
b) 2x2 - 6x
\(=2x\left(x+3\right)\)
c) x2 - y2 - 3x - 3y
\(=\left(x^2-y^2\right)-\left(3x+3y\right)\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-3\right)\)
d) 2x2 - 5x - 3
\(=2x^2-6x+x-3\)
\(=\left(2x^2-6x\right)+\left(x-3\right)\)
\(=2x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(2x+1\right)\)
e) x4 - y4
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
f) mx - my + nx - ny
\(=\left(mx-my\right)++\left(nx-ny\right)\)
\(=m\left(x-y\right)+n\left(x-y\right)\)
\(=\left(x-y\right)\left(m+n\right)\)
TT
17 tháng 11 2017
Bài 1:
a,\(x\left(y+z\right)+3\left(y+z\right)\)
\(=\left(x+3\right)\left(y+z\right)\)
b,\(2x^2-6x\)
\(=2x\left(x-6\right)\)
c,\(x^2-y^2-3x-3y\)
\(=\left(x^2-y^2\right)+\left(-3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x-y-3\right)\left(x+y\right)\)
d,\(2x^2-5x-3\)
\(=2x^2-6x+1x-3\)
\(=\left(2x^2-6x\right)+\left(1x-3\right)\)
\(=2x\left(x-3\right)+1\left(x-3\right)\)
\(=\left(2x+1\right)\left(x-3\right)\)
e,\(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
f,\(mx-my+nx-ny\)
\(=\left(mx-my\right)+\left(nx-ny\right)\)
\(=m\left(x-y\right)+n\left(x-y\right)\)
\(=\left(m+n\right)\left(x-y\right)\)
15 tháng 10 2017
\(a,16x^2-4y^2\)
\(=\left(4x\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y\right)\left(4x+2y\right)\)
\(=\left[2\left(2x-y\right)\right]\left[2\left(2x+y\right)\right]\)
\(=4\left(2x-y\right)\left(2x+y\right)\)
\(b,mx-my-nx+ny+y^2-2xy+x^2\)
\(=\left(mx-my\right)-\left(nx-ny\right)+\left(y^2-2xy+x^2\right)\)
\(=m\left(x-y\right)-n\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(m-n-x+y\right)\)
\(c,\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y+z\right)^3-x^3\right]-\left[y^3+z^3\right]\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+\left(x+y+z\right)x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left[\left(x+y+z\right)^2+x^2+xy+z^2-y^2+yz-z^2\right]\)
\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz+x^2+xy+z^2-y^2+yz-z^2\right)\)
\(=\left(y+z\right)\left(2x^2+z^2+3xy+3yz+2xz\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
\(\left\{{}\begin{matrix}x+my=3\\m^2x+my=2m^2+m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=3\\\left(m^2-1\right)x=2m^2+m-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=3\\x=\dfrac{2m+3}{m+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2m+3}{m+1}\\y=\dfrac{1}{m+1}\end{matrix}\right.\)
\(P=\left(\dfrac{2m+3}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}=\left(2+\dfrac{1}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}\)
\(=4+\dfrac{4}{m+1}+\dfrac{4}{\left(m+1\right)^2}=\left(\dfrac{2}{m+1}+1\right)^2+3\ge3\)
\(P_{min}=3\) khi \(m=-3\)
TN
5 tháng 7 2017
a, \(\left\{{}\begin{matrix}x+my=3m\\mx-y=m^2-2\end{matrix}\right.\)
Tại m = 1 , ta có:
\(\left\{{}\begin{matrix}x+y=3\\x-y=-1\end{matrix}\right.\)
giải hệ ta được:\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
a, mx^2+ my - nx^2-ny
=>(mx^2+my)-(nx^2+ny)
=>m(x^2+y)-n(x^2+y)
=>(x^2+y)(m-n)
TL
Biến đổi được mx2 + tny + nx2 - ny = ( x2 + y ) ( m -n)
HT