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a) \(\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-y-2\right)\left(x+y\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(1.\)
\(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xyz^2\)
\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)
\(=\left(x^2z-xyz\right)\left(x-z\right)\)
\(=xz\left(x-y\right)\left(x-z\right)\)
\(2.\)
\(x^2-\left(a+b\right)xy+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x^2-bxy-axy+aby^2\)
\(=x\left(x-by\right)-ay\left(x-by\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(3.\)
\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)
\(=abx^2+aby^2+a^2xy+b^2xy\)
\(=abx^2+b^2xy+a^2xy+aby^2\)
\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)
\(=\left(ax+by\right)\left(bx+ay\right)\)
\(4.\)
\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)
\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)
\(5.\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)
\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)
\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(6.\)
\(16x^2-40xy+2y^2\)
\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)
\(=\left(4x-5y\right)^2\)
\(7.\)
\(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)
\(=\left(5x^2+y\right)^2\)
\(8.\)
\(-16x^4y^6-24x^5y^5-9x^6y^4\)
\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)
\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)
\(=\left(4x^2y^3+3x^3y^2\right)^2\)
\(9.\)
\(16x^2-4y^2-8x+1\)
\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)
\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)
\(=\left(4x+1\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)
\(10.\)
\(49x^2-25+42xy+9y^2\)
\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)
\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)
\(=\left(7x+3y\right)^2-5^2\)
\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)
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.
1: \(=x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3+y\right)\left(x+3-y\right)\)
2: \(x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-25\)
\(=\left(x-5-y\right)\left(x+5-y\right)\)
4: \(=y\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(y-5\right)\)
5: \(=x^3\left(x+3\right)-9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^3-9\right)\)
Phân tích đa thức thành nhân tử :
a, (x2+y2-5)2 - 4(xy+2)2
b, x2-6x-4y2+12y
c, 9x2-4-y2+4y
d, x2-y24z2-4yz - 10x + 25
Hi hi không biết làm!!!
Câu a sau 4(xy+2) là ^2 nhé mình nhầm TOT