1 . Phan tich da thuc thanh nhan tu :
x^2 + 6x - 9 -y^2
2. Tim x :
x^2 -x - 12 = 0
Phan tich da da thuc thanh nhan phan tu
(x^2+x+1)(x^2+x+2)-12
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
Phan tich da thuc sau thanh nhan tu
6x^3+x^2+x+1
\(6x^3+x^2+x+1=\left(6x^3+3x^2\right)+\left(-2x^2-x\right)+\left(2x+1\right)\)
\(=3x^2.\left(2x+1\right)-x.\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)
K sai dau
giao an truong Tran dai nghia do
Phan tich da thuc thanh nhan tu
( x^2-6x+ 8)( x^2-8x +15) +1
phan tich cac da thuc sau thanh nhan tu:
(y^2+Y)^2 -9y^2 - 9y+20
(X+3)*(x+6)*(x+9)*(x+12)+81
dat y^2+y=z cho gon
\(z^2-9z+20=z^2-4z-5z+20=z\left(z-4\right)-5\left(z-4\right)=\left(z-4\right)\left(z-5\right)\)
\(thaylai:\left(y^2+y-4\right)\left(y^2+y-5\right)\)
tim x bang cach phan tich da thuc thanh nhan tu
x3-3x2+3x-1=0
\(x^3-3x^2+3x-1\) =0
=>\(\left(x-1\right)^3\)=0
=>x-1=0
=>x=1
vậy x =1
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Phan tich da thuc thanh nhan tu
(X^2-6x+8)(x^2-8x+15)+1
phan tich da thuc thanh nhan tu
6x2 - 3xy + x + y -1
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, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
phan tich da thuc sau thanh nhan tu
6x^3+x^2+x+1