phan tich da huc sau thanh nhan tu x2+2xy-8
ai co bai tap ve chuong 1 dai so ko
phan tich da thuc thanh nhan tu
x^2-x-y^2-y
x^2-2xy+y^2-z^2
bai 32 va 33 sbt
lop 8 bai phan tich da thuc thanh nhan tu bang cach nhom hang tu
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
con bai 32, 33 neu ban tra loi duoc minh h them
1+2xy-x2-y2
giải bài tap phan tich da thuc thanh nhan tu
1 + 2xy - x2 - y2
= 1 - x2 + 2xy - y2
= 1 - ( x2 - 2xy + y2 )
= 1 - ( x - y )2
= ( 1 + x - y ).( 1 - x + y )
phan tich da huc thanh nhan tu
3x2-7x-10
\(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
phan tich da thuc thanh nhan tu x2+2xy+7x+7y+10
bai 1 phan tich da thuc sau thanh nhan tu x3 - x2 - x -2
Ta có:
\(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+x-2=\left(x-2\right)\left(x^2+x+1\right)\)
Phan tich da thuc thanh nhan tu : x2 - 4x -y2+4
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^2-4x-y^2+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
Phan tich da thuc thanh nhan tu : x2 - 4x -y2+4
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
phan tich thanh cac nhan tu da thuc
x^2-25+y+2xy
phan tich da thuc thanh nhan tu
9-x^2+2xy-y^2
\(=3^2-\left(x-y\right)^2=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]=\left(3-x+y\right)\left(3+x-y\right)\)
\(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3-x-y\right)\)