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24 tháng 3 2020
a, x2-5xy+2x-10y = (x2 + 2x)-(5xy+10y)
= x(x+2)-5y(x+2)
= (x+2)(x-5y)
b, x2-5x+4 = x2- x - 4x +4
= (x2-x)-(4x-4)
=x(x-1)-4(x-4)
=(x-1)(x-4)
NC
14 tháng 12 2020
\(a,x^2-5xy+2x-10y\)
\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)
\(=x\left(x-5y\right)+2\left(x-5y\right)\)
\(=\left(x-5y\right)\left(x+2\right)\)
\(b,x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=x\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
27 tháng 7 2017
a) \(x^2+2x+1-y^2 = (x^2 +2x+1) – y^2\)
\(= (x+1)^2 - y^2\)
\(= (x+1+ y) (x+1- y)\)
b) \(2xy+z+2x+yz\)
\(=\left(2xy+2x\right)+\left(z+yz\right)\)
\(=2x\left(y+1\right)+z\left(y+1\right)\)
\(=\left(2x+z\right)\left(y+1\right)\)
c) \(x^2+x-xy-y\)
\(=\left(x^2+x\right)-\left(xy+y\right)\)
\(=x\left(x+1\right)-y\left(x+1\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
d) \(5x^2-5xy+10x+10y\)
\(=\left(5x^2-5xy\right)+\left(10x+10y\right)\)
\(=5x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-10\right)\)
27 tháng 7 2017
a, x2+2x+1-y2= (x+1)2-y2= (x+1-y)(x+1+y)
b, 2xy+z+2x+yz= (2xy+2x)+(z+yz)
= 2x(y+1)+z(y+1)
= (2x+z)(y+1)
c, x2 +x - xy - y = x(x+1) - y(x+1)
= (x - y)(x+1)
d, 5x2 - 5xy - 10x + 10y=5x(x-y) - 10(x-y)
= (5x-10)(x-y).(Câu này có thể là sai đề nên mik đã sửa lại đề cho dễ làm nha). Chúc bạn học tốt!!!
TG
16 tháng 8 2020
a) \(x\left(x-y\right)+x-y\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
b) \(2x+2y-x\left(x+y\right)\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x\right)\)
c) \(5x^2-5xy-10x+10y\)
\(=5x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-10\right)\)
d) \(4x^2+8xy-3x-6y\)
\(=4x\left(x+2y\right)-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(4x-3\right)\)
e) \(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2+2y^2-2\right)-\left(x^2z-z+y^2z\right)\)
\(=2\left(x^2+y^2-1\right)-z\left(x^2-1+y^2\right)\)
\(=\left(x^2+y^2-1\right)\left(2-z\right)\)
10 tháng 8 2016
giải hộ câu c, d và f thôi nhá, mấy câu kia biết là rồi
G
31 tháng 10 2020
a) \(6x^3-12x^2y^2+6xy^3=6x.\left(x^2-2xy^2+y^3\right)\)
b) \(\left(x^2+4\right)^2-16=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\)
c) \(5x^2-5xy-10x+10y=\left(5x^2-5xy\right)-\left(10x-10y\right)=5x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-10\right)=5\left(x-y\right)\left(x-2\right)\)
d) \(a^3-3a+3b-b^3=\left(a^3-b^3\right)-\left(3a-3b\right)=\left(a-b\right)\left(a^2+ab+b^2\right)-3.\left(a-b\right)\)
\(=\left(a-b\right)\left(x^2+ab+b^2-3\right)\)
e) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-1-y\right)\left(x-1+y\right)\)
f) \(x^2-x-2=x^2+x-2x-2=\left(x^2+x\right)-\left(2x+2\right)=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
g) \(x^4-5x^2+4=x^4-4x^2+4-x^2=\left(x^4-4x^2+4\right)-x^2=\left(x^2-2\right)^2-x^2\)
\(=\left(x^2-2-x\right)\left(x^2-2+x\right)\)
j) \(x^3-x^3-2x^2-x=-2x^2-x=-\left(2x^2+x\right)=-x\left(2x+1\right)\)
k) \(\left(a^3-27\right)-\left(3-a\right)\left(6a+9\right)=\left(a-3\right).\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\)
\(\left(a-3\right)\left(a^2+3a+9+6a+9\right)=\left(a-3\right)\left(a^2+9a+18\right)\)
h) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)
\(=\left(x^2y-y^2x\right)-\left(x^2z-y^2z\right)+\left(z^2x-z^2y\right)\)
\(=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left[\left(xy-zx\right)-\left(zy-z^2\right)\right]\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)
\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
23 tháng 10 2019
a) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x(x - 2) - 4(x - 2) = (x - 4)(x - 2)
b) x2 - 5x - 14 = x2 - 7x + 2x - 14 = x(x - 7) + 2(x - 7) = (x + 2)(x - 7)
c) 4x2 - 36x + 56 = 4(x2 - 9x + 14) = 4(x2 - 2x - 7x + 14) = 4[x(x - 2) - 7(x - 2)] = 4(x - 7)(x - 2)
d) 3x2 - 16x + 5 = 3x2 - 15x - x + 5 = 3x(x - 5) - (x - 5) = (3x - 1)(x - 5)
f) 8x2 + 30x + 7 = 8x2 + 16x + 14x + 7 = 8x(x + 2) + 7(x + 2) = (8x + 8)(x + 2)
g) 2x2 - 5xy - 12y2 = 2x2 - 8xy + 3xy - 12y2 = 2x(x - 4y) + 3y(x - 4y) = (2x + 3y)(x - 4y)
h) x2 - 7xy + 10y2 = x2 - 2xy - 5xy + 10y2 = x(x - 2y) - 5y(x - 2y) = (x - 5y)(x - 2y)
7 tháng 8 2020
Bài 3 : Ta có : \(A=\frac{2}{5}xy\left(x^2y-5x+10y\right)\)
\(A=\frac{2}{5}xy\cdot x^2y+\frac{2}{5}xy\left(-5x\right)+\frac{2}{5}xy\cdot10y\)
\(A=\frac{2}{5}x^3y^2-2x^2y+4xy^2\)
Chọn C
Bài 4 : \(\left(x-2\right)\left(x+5\right)=x\left(x+5\right)-2\left(x+5\right)\)
\(=x^2+5x-2x-10\)
\(=x^2+3x-10\)
Chọn B
7 tháng 8 2020
Bài 3 :
Ta có: A = 2/5xy( x2y -5x + 10y )
= 2/5xy.x2y - 2/5xy.5x + 2/5xy.10y
= 2/5x3y2 - 2x2y + 4xy2.
Chọn đáp án C
Bài 4 :
Ta có ( x - 2 )( x + 5 )
= x( x + 5 ) - 2( x + 5 )
= x2 + 5x - 2x - 10 = x2 + 3x - 10.
Chọn đáp án B.
Hok tốt
KB
10 tháng 9 2018
1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)
2 ) \(2x\left(b-a\right)+a-b=2x\left(b-a\right)-\left(b-a\right)=\left(2x-1\right)\left(b-a\right)\)
3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)
4 ) \(x^2-xy-2x+2y=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
5 ) \(5x^2y+5xy^2+a^2x+a^2y\)
\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)
\(=\left(5xy+a^2\right)\left(x+y\right)\)
6 ) \(2x^2-6xy+5x-15y\)
\(=2x\left(x-3y\right)+5\left(x-3y\right)\)
\(=\left(2x+5\right)\left(x-3y\right)\)
7 ) \(ax^2-3axy+bx-3by\)
\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)
\(=x\left(ax+b\right)-3y\left(ax+b\right)\)
\(=\left(x-3y\right)\left(ax+b\right)\)
8 ) \(x^2+4x-5x-20=0\)
\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
9 ) \(x^2+10x-2x-20=0\)
\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
10 ) \(x^2-6x-4x+24=0\)
\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
:D
a) \(x^2-5xy+2x-10y\)
\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)
\(=x.\left(x-5y\right)+2.\left(x-5y\right)\)
\(=\left(x-5y\right).\left(x+2\right)\)
b) \(x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=\left(x^2-4x\right)-\left(x-4\right)\)
\(=x.\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-4\right).\left(x-1\right)\)
Chúc bạn học tốt!