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Sửa đề \(x^2-3xy-10y^2\)
\(=x^2-5xy+2xy-10y^2\)
\(=x\left(x-5y\right)+2y\left(x-5y\right)=\left(x-5y\right)\left(x+2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2-3xy-10y2
=x2+2xy-5xy-10y2
=x(x+2y)-5y(x+2y)
=(x-5y)(x+2y)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( -5x2 +3xy + 7) + ( -6x2y + 4xy2 - 5)=4*x*y^2-6*x^2*y+3*a*x*y-5*a*x^2+7*a-5
b) ( 2,4x3 - 10x2y) + (7x2y - 2,4x3 + 3xy2)=3*x*y^2-3*x^2*y
c) ( 15x2y - 7xy2 - 6y2) + (2x2 - 12x2y + 7xy2)=-6*y^2+3*x^2*y+2*x^2
d) ( 4x2 + x2y - 5y3) + (5/3 x3 - 6xy2 - x2y) + (x3/3 + 10y3) + ( 6y3-15xy2 - 4x2y - 10x3)=11*y^3-21*x*y^2-4*x^2*y-8*x^3+4*x^2
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích đa thức thành nhân tử:
a) 5x - 5y
= 5(x - y)
b) 3xy2 + x2y
= xy(3y + x)
c) 12x2y - 18xy2 - 30y3
= 2y(6x2 - 9xy - 15y2)
= 2y(6x2 + 6xy - 15xy - 15y2)
= 2y[6x(x + y) - 15y(x + y)]
= 2y(x + y)(6x - 15y)
= 6y(x + y)(2x - 5y)
d) -17x3y - 34x2y2 + 51xy3
= -17xy(x2 + 2xy - 3y2)
= -17xy(x2 - xy + 3xy - 3y2)
= -17xy[x(x - y) + 3y(x - y)]
= -17xy(x - y)(x + 3y)
e) x(y - 1) + 3(y - 1)
= (y - 1)(x + 3)
f) 162(x - y) - 10y(y - x)
= 162(x - y) + 10y(x - y)
= (x - y)(162 + 10y)
= (x - y)(256 + 10y)
a) Ta có: 5x-5y
=5(x-y)
b) Ta có: \(3xy^2+x^2y\)
\(=xy\left(3y+x\right)\)
c) Ta có: \(12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left(2x^2-5xy+2xy-5y^2\right)\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
d) Ta có: \(-17x^3y-34x^2y^2+51xy^3\)
\(=-17xy\left(x^2+2xy-3y^2\right)\)
\(=-17xy\left(x^2+3xy-xy-3y^2\right)\)
\(=-17xy\left[x\left(x+3y\right)-y\left(x+3y\right)\right]\)
\(=-17xy\left(x+3y\right)\left(x-y\right)\)
e) Ta có: x(y-1)+3(y-1)
=(y-1)(x+3)
![](https://rs.olm.vn/images/avt/0.png?1311)
giải hộ câu c, d và f thôi nhá, mấy câu kia biết là rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x^2+4x=x\left(x+4\right)\)
b, \(16x^2-9=\left(4x-3\right)\left(4x+3\right)\)
c, \(2x-2y+bx-by=2\left(x-y\right)+b\left(x-y\right)\)
\(=\left(b+2\right)\left(x-y\right)\)
d, \(x^2+4x-5=x^2+5x-x-5\)
\(=x\left(x+5\right)-\left(x+5\right)=\left(x-1\right)\left(x+5\right)\)
e, \(x^3+x+5x^2+5=x\left(x^2+1\right)+5\left(x^2+1\right)\)
\(=\left(x+5\right)\left(x^2+1\right)\)
f, \(x^2+2xy-9+y^2\)
\(=\left(x+y\right)^2-9\)
\(=\left(x+y+3\right)\left(x+y-3\right)\)
g, \(x^2-3xy-10y^2=x^2+2xy-5xy-10y^2\)
\(=x\left(x+2y\right)-5y\left(x+2y\right)\)
\(=\left(x-5y\right)\left(x+2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
\(x^2-3xy-10y^2=x^2-5xy+2xy-10y^2=x\left(x-5y\right)+2y\left(x-5y\right)=\left(x-5y\right)\left(x+2y\right)\)