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b mk thấy nó sai đề sao ý
c) \(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)
\(=\left(x^2+x+4\right)^2+2.4x.\left(x^2+x+4\right)+16x^2-x^2\)
\(=\left(x^2+x+4+4x\right)^2-x^2\)
\(=\left(x^2+5x+4\right)^2-x^2\)
\(=\left(x^2+5x+4-x\right)\left(x^2+5x+4+x\right)=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
b. \(\left(a^2+a\right)+a\left(a^2+a\right)-12\)
<=>\(\left(x^3+3x^2-4\right)+\left(3x^2+9x-12\right)\)
<=>\(x\left(x^2+3x-4\right)+3\left(x^2+3x-4\right)\)
<=>\(\left(x^2+3x-4\right)\left(x+3\right)\)
<=>\(\left(x+3\right)\left(x^2+4x\right)-\left(x-4\right)\)
đóngmở ngoặc nhé mk ngại ghi lại
<=>(x+3)(x(x+4)-(x+4))
<=>(x+3)(x-1)(x+4)
kết pn fb mk nhé longtrangv@gmail.com
c) \(x^3-x^2-4x^2+8x-4\)
= \(x^3-x^2-4x^2+4x+4x-4\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
a) x2 – 4 + (x – 2)2
= (x2 – 22) + (x – 2)2 = (x – 2)(x + 2) + (x – 2)2
= (x – 2) [(x + 2) + (x – 2)]
= (x – 2)(x + 2 + x – 2)
= 2x(x – 2)
b) x3 – 2x2 + x – xy2
= x(x2 – 2x + 1 – y2) = x[(x2 – 2x + 1) – y2]
= x[(x – 1)2 – y2]
= x[(x – 1) + y] [(x – 1) – y]
= x(x – 1 + y)(x – 1 – y)
c) x3 – 4x2 – 12x + 27
= (x3 + 27) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9 – 4x)
= (x + 3)(x2 – 7x + 9)
a, \(x^6-x^4-9x^3+9x^2\)
= \(x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
=\(x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
= \(\left(x-1\right)\left(x^4\left(x+1\right)-9x^2\right)\)
= \(\left(x-1\right)\left(x^5+x-9x^2\right)\)
b, \(x^4-4x^3+8x^2-16x+16\)
= \(x^4-4x^3+4x^2+4x^2-16x+16\)
\(=x^2\left(x^2-4x+4\right)+4\left(x^2-4x+4\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
c, \(\left(xy+4\right)^2-4\left(x+y\right)^2\)
= \(\left(xy+4\right)^2-\left(2\left(x+y\right)\right)^2\)
= \(\left(xy-2x-2y+4\right)\left(xy+2x+2y+4\right)\)
= \(\left(x\left(y-2\right)-2\left(y-2\right)\right)\left(x\left(y+2\right)+2\left(y+2\right)\right)\)
=\(\left(x-2\right)\left(y-2\right)\left(x+2\right)\left(y+2\right)\)
d, \(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2\)
= \(a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2-2ab+2ac-2bc-4b^2\)
=\(2a^2+2b^2+2c^2+4ac-4b^2\)
Lời giải:
a.
$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$
$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$
b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?