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Ta có ; x2 - 11x + 24
= x2 - 3x - 8x + 24
= x(x - 3) - (8x - 24)
= x(x - 3) - 8(x - 3)
= (x - 3)(x - 8)
a )\(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
b )\(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+2x+2\right)\left(x^2-2\right)\)
c ) \(x^2\left(1-x^2\right)-4-4x^2=x^2-x^4-4-4x^2\)
\(=x^2-\left(x^2+2\right)^2=\left(x-x^2-2\right)\left(x^2+x+2\right)\)
b, \(\left(x^2+x\right)^2+4x^2+4x-12=x^4+2x^3+x^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x^3+3x^2+8x+12\right)\left(x-1\right)\)
\(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\left(x-1\right)\)
\(=\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
c, \(x^3+3x^2-4=\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(2x+4\right)\)
\(=x^2\left(x+2\right)+x\left(x+2\right)-2\left(x+2\right)\)
= \(\left(x^2+x-2\right)\left(x+2\right)\)
a)\(x^5+x^4+1=x^5-\left(-x^3+x^3\right)+x^4+\left(x^2-x^2\right)+\left(x-x\right)+1\)
\(=x^5-x^3+x^2+x^4-x^2+x+x^3-x+1\)
\(=x^2\left(x^3-x+1\right)+x\left(x^3-x+1\right)+\left(x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
b,c có ng lm rồi
d)\(2x^4-3x^3-7x^2+6x+8\)
Ta thấy x=-1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x+1
\(\Rightarrow\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)
\(\Rightarrow\left(x+1\right)\left[2x^3-x^2-4x-4x^2+2x+8\right]\)
\(\Rightarrow\left(x+1\right)\left[x\left(2x^2-x-4\right)-2\left(2x^2-x-4\right)\right]\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
phần còn lại bạn tự lo nhé
b) \(x^3-3x^2+2\)
\(=x^3-2x^2-x^2+2\)
\(=x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x^2-x-2\right)\left(x-2\right)\)
c) \(x^4y^4+64\)
\(=x^4y^4+16x^2+64-16x^2\)
\(=\left(x^2y^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2y^2-4x+8\right)\left(x^2y^2+4x+8\right)\)
d) \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+1\)
\(=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^6-x^4-x+x^3-1\right]\)
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) \(12x-9-4x^2\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left(2x-3\right)^2\)
b)\(1-9x+27x^2-27x^3\)
\(=\left(1-3x\right)^{^3}\)
c)\(\frac{x^2}{4}+2xy+4y^2-25\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
d)\(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)^2-3\left(x^2-4x\right)-5\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)\left(x^2-4x-3\right)-5\left(x^2-4x-3\right)\)
\(=\left(x^2-4x-5\right)\left(x^2-4x-3\right)\)
\(=\left(x^2+x-5x-5\right)\left(x^2-4x-3\right)\)
\(=\left[x\left(x+1\right)-5\left(x+1\right)\right]\left(x^2-4x-3\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x^2-4x-3\right)\)
Chúc bạn học tốt !
a) x^2+4xy-16+4y^2
=(x^2+4xy+4y^2)-4^2
=(x+2y)^2-4^2
=(x+2y-4)(x+2y+4)
b)27-(x-1)^3
=3^3-(x-1)^3
=(4-x)(5+4x)
c)x^2-4x+3
=x^2-x-3x+3
=x(x-1)-3(x-1)
=(x-1)(x-3)
d) x^2-x-12
=x^2-4x+3x-12
=x(x-4)+3(x-4)
=(x-4)(x+3)
e) x^4+4
=(x^2)^2+2x^2.2+2^2-2x^2.2
=(x^2+2)^2-4x^2
=(x^2-2x+2)(x^2+2x+2)
tick nha