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A = x2(x - 1) + 6(1 - x)
A = x3 - x2 + 6 - 6x
A = (x3 - 6x) - (x2 - 6)
A = x.(x2 - 6) - (x2 - 6)
A = (x - 1)(x2 - 6)
C = x2 + 2xy + y2 - yz - xz
C = (x + y)2 - z.(x + y)
C = (x + y - z).(x + y)
a) \(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+x-3=\left(x-3\right)\left(1-5x\right).\)
b) \(x^4+64=x^4+16x^2+64-16x^2=\left(x^2+8\right)^2-\left(4x\right)^2=\left(x^2+4x+8\right)\left(x^2-4x+8\right).\)
c) \(64x^2+4y^4=4\left(16x^2+y^4\right)\)
d) \(x^5+x-1\)đa thức này có nghiệm vô tỷ. Mik ko phân tích được.
a/ \(x^4+16\)
\(=x^4+4x^2+16-4x^2\)
\(=\left(x^4+4x^2+16\right)-4x^2\)
\(=\left(x^2+4\right)^2-\left(2x\right)^2\)
\(=\left(x^2+4-2x\right)\left(x^2+4+2x\right)\)
b/ \(64x^4+y^4\)
\(=64x^4+y^4+16x^2y^2-16x^2y^2\)
\(=\left(64x^4+y^4+16x^2y^2\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(y^2+8x^2-4xy\right)\left(8x^2+y^2-4xy\right)\)
\(a,x^4+64=\left(x^4+16x^2+64\right)\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right).\left(x^2+4x+8\right)\)
\(b,x^5+x+1\)
\(=\left(x^2+x+1\right).\left(x^3-x^2+1\right)\)
...
y^4+64
=(y^2)^2+16y^2+64-16y^2
=(y^2+8-4x)(x^2+8+4x)
x^2+4
=x^2+2x^2+4-2x^2
=(x+2)^2-2x^2
=(x^2+2-2x)(x^2+2+2x)
x^4+16
=(x^2)^2+4x^2+16-4x^2
=(x+4)^2-4x^2
=(x^2+4-4x)(x^2+4+4x)
x^4y^4+4
=x^4y^4+4x^4+2^2-4x^4
=(x^4y^4+2)^2-(2x^2)^2
=(x^4y^4+2+2x^2)(x^4y^4+2-2x^2)
4x^4y^4+1
=4x^4y^4+x^4+1-x^4
=(2x^4y^4+1)^2-(x^2)^2
=(2x^4y^4+1-x^2)(2x^4y^4+1+x^2)
Mình ko bt câu D đúng hay sai nữa. Mà lỡ sai bạn đừng giận mình nha!
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
a/ Xem lại đề
b/ \(\left(y-x\right)\left(y+x\right)\left(2y-x\right)\left(2y+x\right)\)
c/ \(\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
d/ \(\left(2x^2-5x+1\right)\left(2x^2+5x+1\right)\)
a) Ta có: \(x^4+64\)
\(=x^4+16x^2+64-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
b) Ta có: \(81x^4+4y^4\)
\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)
\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)
\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)
c) Ta có: \(x^5+x+1\)
\(=x^5+x^2-x^2+x-1\)
\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)