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a: x^2+4xy-21y^2
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
c: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
=x(x+5y)-3y(x+5y)
=(x+5y)(x-3y)
d: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
a) \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
c) \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
Bài 3:
a) Ta có: \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) Ta có: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(x+y\right)\left(5x+y\right)\)
c) Ta có: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) Ta có: \(\left(x-y\right)^2+4\left(x-y\right)-12\)
\(=\left(x-y\right)^2+6\left(x-y\right)-2\left(x-y\right)-12\)
\(=\left(x-y\right)\left(x-y+6\right)-2\left(x-y+6\right)\)
\(=\left(x-y+6\right)\left(x-y-2\right)\)
e) Ta có: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5y\right)\)
f) Ta có: \(x^2yz+5xyz-14yz\)
\(=yz\left(x^2+5x-14\right)\)
\(=yz\left(x^2+7x-2x-14\right)\)
\(=yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\)
\(=yz\left(x+7\right)\left(x-2\right)\)
Ta có : 5x2 + 6xy + y2
= 5x2 + 5xy + xy + y2
= 5x(x + y) + y(x + y)
= (5x + y)(x + y)
Cách tách hạng tử như sau:
Cho đa thức \(ax^2+bx+c\)
Ta tách hạng tử \(bx=mx+nx\)sao cho \(m.n=a.c\)
Sau đó gộp lại ta được \(\left(ax^2+mx\right)+\left(nx+c\right)\)
Tiếp túc đặt nhân tử chung ta được một tích.
Trên đây là cách tách hạng tử, bạn áp dụng vào làm nhé!
Ta có : x2 + 2xy - 15y2
= x2 - 3xy + 5xy - 15y2
= x(x - 3y) + 5y(x - 3y)
= (x - 3y)(x + 5y)
X2+4xy-21y2=(x2+4xy+4y2)-25y2=(x+2)2-(5y)2=(x+2-5y)(x+2+5y)
5x2+6xy+y2=9x2+6xy+y2-4x2=(3x+y)2-4x2=(3x+y+2x)(3x+y-2x)
(x-y)2+4(x-y)-12=(x-y+2)2-16=(x-y+2+4)(x-y+2-4)
x2-7xy+10y2=x2-7xy+\(\frac{49y^2}{4}-\frac{9y^2}{4}\)= \(\left(x-\frac{7}{2}\right)^2-\left(\frac{3y}{2}\right)^2\)=\(\left(x-\frac{7}{2}-\frac{3y}{2}\right)\left(x-\frac{7}{2}+\frac{3y}{2}\right)\)
x2+2xy-15y2=(x+y)2-16y2=(x+y-4y)(x+y+4y
2) \(5x^2+6xy+y^2\)
\(=9x^2+6xy+y^2-4x^2\)
\(=\left(3x+y\right)^2-\left(2x\right)^2\)
\(=\left(3x+y+2x\right)\left(3x+y-2x\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
3) \(x^2+2xy-15y^2=x^2+2xy+y^2-16y^2\)
\(=\left(x+y\right)^2-\left(4y\right)^2\)
\(=\left(x+y+4y\right)\left(x+y-4y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
hc tốt
f)\(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)
h)\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)
g)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
f)\(x^2-5x-14=x^2-7x+2x-14\)
\(=\left(x+2\right)\left(x-7\right)\)
i)\(x^2-7x+10=x^2-5x-2x+10\)
\(=\left(x-2\right)\left(x-5\right)\)
h)\(x^2-7x+12=x^2-4x-3x+12\)
\(=\left(x-3\right)\left(x-4\right)\)
g)\(x^2+6x+5=x^2+x+5x+5\)
\(=\left(x+5\right)\left(x+1\right)\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
\(a,x^2+4xy-21y^2\\ =x^2+7xy-3xy-21y^2\\ =x\left(x+7y\right)-3y\left(x+7y\right)\\ =\left(x+7y\right)\left(x-3y\right)\\ b,5x^2+6xy+y^2\\ =5x^2+5xy+xy+y^2\\ =5x\left(x+y\right)+y\left(x+y\right)\\ =\left(x+y\right)\left(5x+y\right)\\ c.x^2+2xy-15y^2\\ =x^2+5xy-3xy-15y^2\\ =x\left(x+5y\right)-3y\left(x+5y\right)\\ =\left(x+5y\right)\left(x-3y\right)\)
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