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CM
25 tháng 12 2018
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
MS
1
LN
10 tháng 10 2021
\(a,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right).\left(x-y+z\right)\)
\(b,x^3+y^3+2x^2-2xy+2y^2=\left(x^3+y^3\right)+2\left(x^2-xy+y^2\right)=\left(x+y\right).\left(x^2-2xy+y^2\right)+2.\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right).\left(x+y+2\right)\)
PB
1
21 tháng 10 2021
1: \(-x^2+2x+8\)
\(=-\left(x^2-2x-8\right)\)
\(=-\left(x-4\right)\left(x+2\right)\)
2: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)
28 tháng 10 2021
\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)
31 tháng 7 2021
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
31 tháng 7 2021
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
3 tháng 9 2021
e: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
f: \(x^2-2x+7x-14\)
\(=x\left(x-2\right)+7\left(x-2\right)\)
=(x-2)(x+7)
h: \(5x^2-10xy+5y^2-20\)
\(=5\left(x^2-2xy+y^2-4\right)\)
\(=5\left(x-y-2\right)\left(x-y+2\right)\)
3 tháng 9 2021
a: \(3x^4-6x^3+2x^2=x^2\left(3x^2-6x+2\right)\)
b: \(x^3y+12x^2y+36xy=xy\left(x^2+12x+36\right)=xy\left(x+6\right)^2\)
c: \(x^3y-9xy^3=xy\left(x^2-9y^2\right)=xy\left(x-3y\right)\left(x+3y\right)\)
d: \(x^2y^2-2xy^2+y^2=y^2\left(x-1\right)^2\)
NH
2
27 tháng 8 2021
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
27 tháng 8 2021
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
a) \(x^2-2xy-3y^2\) \(=x^2-3xy+xy-3y^2=x\left(x-3y\right)+y\left(x-3y\right)\)
\(=\left(x+y\right)\left(x-3y\right)\)
b) \(2x^2-3x-2=2x^2-4x+x-2\) \(=2x\left(x-2\right)+\left(x-2\right)\) \(=\left(x-2\right)\left(2x+1\right)\)