Phan tinh da thuc thanh nhan tu
A, xy † xz — 2y —2z
B, x^2-—6xy †9y^2 —25z^2
Phan tich da thuc thanh nhan tu
P=x^2 - 6xy +9y^2
\(P=x^2-6xy+9y^2=\left(x-3y\right)^2\)
(Áp dụng 7 hằng đẳng thức đáng nhớ)
1.phan tich cac da thưc sau thành nhan tử
a. xy + xz -2y -2z
b. x2- 6xy + 9y2 - 25z2
c. x2 -2x + 2y -xy
d. (x2 + 1)2 - 4x2
e. x2 -y2 +2yz - z2
mong mọi người jup dỡ cảm ơn nhiều ạ.
a) xy+xz-2y-2z=x(y+z)-2(y+z)=(y+z)(x-2)
b) \(x^2-6xy+9y^2-25z^2=\left(x-3y\right)^2-\left(5z\right)^2=\left(x-3y-5z\right)\left(x-3y+5z\right)\)
c) \(x^2-2x+2y-xy=x\left(x-2\right)+y\left(2-x\right)=x\left(x-2\right)-y\left(x-2\right)=\left(x-2\right)\left(x-y\right)\)
d) \(\left(x^2+1\right)^2-4x^2=\left(x^2+1-2x\right)\left(x^2+1+2x\right)=\left(x-1\right)^2\left(x+1\right)^2\)
e)\(x^2-y^2+2yz-z^2=x^2-\left(y^2-2yz+z^2\right)=x^2-\left(y-z\right)^2=\left(x+y-z\right)\left(x-y+z\right)\)
phan tich da thuc thanh nhan tu:
x^2-xy-2y^2
x2-xy-2y2
= x2-2xy+xy-2y2
=x(x-2y)+y(x-2y)
=(x-2y)(x+y)
Phan tich da thuc thanh nhan tu:
P=x^2-6xy+9y^2
Cac ban ghi loi giai ro rang từng bước cho mình nha
\(x^2-6xy+9y^2=x^2-2.\left[x.\left(3y\right)\right]+\left(3y\right)^2\)
\(=\left(x-3y\right)^2\)
Phan tich da thuc thanh nhan tu
1)P= x2 - 6xy+9y2
2) P= x4 - 64x
1)P= x2 - 6xy+9y2
= (x-3y)^2
2) P= x4 - 64x
= (x^3 - 64) x
= (x^3 - 4^3)x
= (x^3 - 12x^2 + 48x - 64)x
m.n giúp e với ạ :<
1) xy + xz - 2y - 2z
2) x^2 - 6xy + 9y^2 - 25z^2
3) 3x^2 - 3y^2 - 12x + 12y
e cảm ơnnn
Lời giải:
1. $xy+xz-2y-2z=(xy+xz)-(2y+2z)$
$=x(y+z)-2(y+z)=(x-2)(y+z)$
2. $x^2-6xy+9y^2)-25z^2$
$=(x-3y)^2-(5z)^2=(x-3y-5z)(x-3y+5z)$
3.
$3x^2-3y^2-12x+12y=(3x^2-3y^2)-(12x-12y)=3(x^2-y^2)-12(x-y)=3(x-y)(x+y)-12(x-y)=3(x-y)(x+y-4)$
phan tich da thuc thanh nhan tu :xy(x-y)-xz(x+z)+yz(2x+z-y)
\(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz=xy\left(x+y+z\right)-xyz+\left(yz+xz\right)\left(x+y+z\right)\)
\(=xy\left(x+y+z-z\right)+z\left(x+y\right)\left(x+y+z\right)\)
\(=xy\left(x+y\right)+z\left(x+y\right)\left(x+y+z\right)\)
\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
phan tich da thuc thanh nhan tu xy+xz+3+3y