Phan tich da thuc thanh nhan tu:
+)\(2x-1-x^2\)
+) \(8x^3+y^6\)
+) \(x^2-16+4xy+4y^2\)
cau a, 2x-2y=
cau b, x2-2x+1=
cau c,y3+4xy2+4y
cau d,x3-8
cau e,x2-2
phan tich cac da thuc tren thanh nhan tu
a)2x-2y=2(x-y)
b)x2-2x+1=(x-1)2
c)y3+4xy2+4y=y(y2+4xy+4)
d)x3-8=x3-23=(x-2)(x2+2x+4)
e)\(x^2-2=x^2-\left(\sqrt{2}\right)^2=\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\)
phan tich Da thuc thanh nhan tu a) x^2-8x+16-y^2 b)x^2+9x+18 c) x^3-7x-6
b) x^2 + 9x + 18
= x^2 + 3x + 6x + 18
= (x^2 + 3x) + (6x + 18)
= x(x + 3) + 6(x + 3)
= (x + 3) (x + 6)
phan tich da thuc sau thanh nhan tu: 3(x+5)(x+6)(x+7)-8x(2 cach)
Phan tich da thuc thanh nhan tu
3x2 -11x+ 68x2+10x-34x2+8x-5x4+4y4(x2-2x)(x2-2x-1)-6Phan tich da thuc thanh nhan tu:
1)P= 9(x+3)^2 - 4(x-2)^2
2) P= 25(2x-y)^2 - 16(x+2y)^2
1)P\(=9\left(x+3\right)^2-4\left(x-2\right)^2\)\(=\left(3x+9\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+9+2x-4\right)\left(3x+9-2x+4\right)\)(hằng đẳng thức số 3)
\(=\left(5x+5\right)\left(x+13\right)\)
\(=5\left(x+1\right)\left(x+13\right)\)
2)P\(=25\left(2x-y\right)^2-16\left(x+2y\right)^2\)\(=\left(10x-5y\right)^2-\left(4x+8y\right)^2\)
\(=\left(14x+3y\right)\left(6x-13y\right)\)(tương tự câu 1)
2x2y-4xy2+6xy
phan tich da thuc thanh nhan tu
\(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)
\(2x^2y-4xy^2+6xy=2xy\cdot\left(x-2y+3\right)\)
x^2-4+4xy-8y. Phan tich da thuc thanh nhan tu
x^2-4+4xy-8y=x^2+4xy+4y^2-4y^2-8y-4=(x+2y)^2-(2y+2)^2=(x+2y-2y+2)(x+2y+2y-2)=(x+2)(x+4y-2)
Phan tich da thuc x^2 + y^3 + 2x^2 -2cy + 2y^2 thanh nhan tu
phan tich da thuc thanh nhan tu
x^2+6x+9
10x-25-x^2
8x^3-1/8
8x^3+12x^2+6xy^2+y^3
\(a,x^2+6x+9\)
\(=\left(x+3\right)^2\)
\(b,10x-25-x^2\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
\(c,8x^3-\frac{1}{8}\)
\(=8x^3-\left(\frac{1}{2}\right)^3\)
\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)
\(d,8x^3+12x^2+6xy^2+y^3\)
\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)
hok tốt!
Điệp viên 007 sai c
c, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)