phan tich da thuc thanh nhan tu
x^2-y^2+2x+1
Phan tich da thuc thanh nhan tu (1+2x)(1-2x)-x(x+2)(x-2)
\(\left(1+2x\right).\left(1-2x\right)-x.\left(x+2\right).\left(x-2\right)\))
\(=1-\left(2x\right)^2-x.x^2-2^2\)
\(=1-4x^2-x^3-4\)
Ko bt có đúng ko nữa
( 1 + 2x ) ( 1 - 2x ) - x ( x + 2 ) ( x - 2 )
= 1 - 4x2 - x ( x2 - 4 )
= 1 - 4x2 - x3 + 4x
= - ( x3 + 4x2 - 4x - 1 )
= - ( x3 - x2 + 5x2 - 5x + x - 1 )
= - [ x2 ( x - 1 ) + 5x ( x - 1 ) + ( x - 1 ) ]
= - ( x - 1 ) ( x2 + 5x + 1 )
phan tich da thuc sau thanh nhan tu:
a)(x-y+4)^2-(2x+3y-1)^2
Đặt \(A=\left(x-y+4\right)^2-\left(3x+3y-1\right)^2\)
Ta có:
\(\left(x-y+4\right)^2=x^2-xy+4x-yx+y^2-4y+4x-4y+16\)
\(=x^2+y^2-2xy+8x-8y+16\)
\(\left(3x+3y-1\right)^2=9x^2+9xy-3x+9xy+9y^2-3y-3x-3y+1\)
\(=9x^2+9y^2-6x-6y+18xy+1\)
Mình làm đến đây bạn trừ 2 kết quả cho nhau rồi sẽ ra
Phan tich da thuc x^2 + y^3 + 2x^2 -2cy + 2y^2 thanh nhan tu
phan tich da thuc thanh nhan tu: x(x+2)(x^2+2x+2)+1
x(x+2)(x^2+2x+2)+1 = (x^2+2x)(x^2+2x+1)+1
Đặt x^2+2x+1=y ta được:
(y-)(y+1)+1=y^2-1+1=y^2
= (x^2+2x+1)^2
= ( x + 1 )^4
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
phan tich da thuc thanh nhan tu :x^5+2x^4+3x^3+2x^2+2x+1
x^5+2x^4+2x^3+2x^2+2x+1
=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)
=x^4(x+1)+x^3(x+1)+x^2(x+1)+x(x+1)+(x+1)
=(x+1)(x^4+x^3+x^2+x+1)
phan tich da thuc thanh nhan tu (x^2-x+2)^2- 2x^2 +2x -7
phan tich da thuc thanh nhan tu C=x^4+2x^3.y-2X^2.y^2+11.x.y^3-6y^4
phan tich da thuc thanh nhan tu (x^2+2x+3).(2x^2+2x+5)-8