\(6x^2-xy-y^2\)

b, \(8x...">

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19 tháng 8 2017

a, \(6x^2-xy-y^2\)

\(=6x^2-3xy+2xy-y^2\)

\(=3x\left(2x-y\right)+2y\left(x-y\right)\)

\(=\left(3x+2y\right)\left(x-y\right)\)

b, \(8x^2-23x-3\)

\(=8x^2-24x+x-3\)

\(=8x\left(x-3\right)+\left(x-3\right)=\left(8x+1\right)\left(x-3\right)\)

c, \(10x^2-11x-6\)

\(=10x^2-15x+4x-6\)

\(=5x\left(2x-3\right)+2\left(2x-3\right)\)

\(=\left(5x+2\right)\left(2x-3\right)\)

d, \(x^3-6x^2+11x-6\)

\(=x^3-3x^2-3x^2+9x+2x-6\)

\(=x^2\left(x-3\right)-3x\left(x-3\right)+2\left(x-3\right)\)

\(=\left(x^2-3x+2\right)\left(x-3\right)\)

\(=\left(x^2-2x-x+2\right)\left(x-3\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

22 tháng 6 2017

Mai cho bn đấy tui dg định off =))

a)\(11x+11y-x^2-xy\)

\(=\left(11x+11y\right)-\left(x^2+xy\right)\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

b)\(x^2-xy-8x+8y\)

\(=\left(x^2-xy\right)-\left(8x-8y\right)\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-8\right)\left(x-y\right)\)

c)\(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

\(=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)

d)\(x^2+2xy+y^2-xz-yz\)

\(=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)

\(=\left(x+y\right)^2-z\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y-z\right)\)

22 tháng 6 2017

a) \(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(x+y\right)\left(11-x\right)\)

b) \(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-y\right)\left(x-8\right)\)

c) \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

\(=\left(x-3\right)^2-y^2\)

\(=\left(x-3-y\right)\left(x-3+y\right)\)

d) \(x^2+2xy+y^2-xz-yz\)

\(=\left(x+y\right)^2-z\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y-z\right)\)

AH
Akai Haruma
Giáo viên
15 tháng 7 2018

a)

\(x^3+6x^2+11x+6=(x^3-x)+(6x^2+12x+6)\)

\(=x(x^2-1)+5(x^2+2x+1)\)

\(=x(x-1)(x+1)+6(x+1)^2\)

\(=(x+1)[x(x-1)+6(x+1)]=(x+1)(x^2+5x+6)\)

\(=(x+1)(x^2+2x+3x+6)\)

\(=(x+1)[x(x+2)+3(x+2)]=(x+1)(x+2)(x+3)\)

b) \(x^3+6x^2-13x-42\)

\(=x^3+2x^2+4x^2+8x-21x-42\)

\(=x^2(x+2)+4x(x+2)-21(x+2)\)

\(=(x+2)(x^2+4x-21)\)

\(=(x+2)[x^2-3x+7x-21)\)

\(=(x+2)(x+7)(x-3)\)

AH
Akai Haruma
Giáo viên
15 tháng 7 2018

c)

\(x^3-5x^2+8x-4=(x^3-x^2)-4x^2+8x-4\)

\(=x^2(x-1)-4(x^2-2x+1)\)

\(=x^2(x-1)-4(x-1)^2\)

\(=(x-1)[x^2-4(x-1)]=(x-1)(x^2-4x+4)\)

\(=(x-1)(x-2)^2\)

d) \(2x^3-x^2+3x+6\)

\(=2x^3+2x^2-3x^2+3x+6\)

\(=2x^2(x+1)-3(x^2-x-2)\)

\(=2x^2(x+1)-3[x^2+x-2x-2]\)

\(=2x^2(x+1)-3[x(x+1)-2(x+1)]\)

\(=2x^2(x+1)-3(x+1)(x-2)\)

\(=(x+1)(2x^2-3x+6)\)

23 tháng 12 2016

a, \(x^4+6x^3+7x^2-6x+1\)

\(=x^4-2x^2+1+6x^3+9x^2+6x\)

\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)

\(=\left(x^2-1+3x\right)^2\)

b, \(x^4-7x^3+14x^2-7x+1\)

\(=x^4+2x^2+1+7x^3+12x^2-7x\)

\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)

\(=\left(x^2-1+3x\right)^2\)

c, \(12x^2-11x-36\)

\(=12x^2-27x+16x-36\)

\(=3x\left(4x-9\right)+4\left(4x-9\right)\)

\(=\left(4x-9\right)\left(3x+4\right)\)

AH
Akai Haruma
Giáo viên
17 tháng 10 2018

a)

\(25x^2-9(x+y)^2=(5x)^2-(3x+3y)^2\)

\(=(5x-3x-3y)(5x+3x+3y)=(2x-3y)(8x+3y)\)

b)

\(x^2-x-2=x^2+x-2x-2=x(x+1)-2(x+1)=(x-2)(x+1)\)

c)

\(3x^2-11x+6=3x^2-9x-2x+6\)

\(=3x(x-3)-2(x-3)=(x-3)(3x-2)\)

d)

\(x^2+5x+8\): biểu thức không phân tích được thành nhân tử

AH
Akai Haruma
Giáo viên
17 tháng 10 2018

e)

\(x^2+8x+7=x^2+x+7x+7\)

\(=x(x+1)+7(x+1)=(x+1)(x+7)\)

g)

\(x^2-6x-16=x^2-6x+9-25\)

\(=(x-3)^2-5^2=(x-3-5)(x-2+5)=(x-8)(x+2)\)

h)

\(4x^2-8x+3=4(x^2-2x+1)-1\)

\(=4(x-1)^2-1=(2x-2)^2-1^2=(2x-2-1)(2x-2+1)\)

\(=(2x-3)(2x-1)\)

i)

\(3x^2-11x+6=3x^2-9x-2x+6\)

\(=3x(x-3)-2(x-3)=(3x-2)(x-3)\)

3 tháng 6 2017

a)\(x^2+9x+20\)

\(\Leftrightarrow x^2+4x+5x+20\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)

\(\Leftrightarrow\left(x+5\right)\left(x+4\right)\)

b)\(x^2+x-12\)

\(\Leftrightarrow x-3x+4x-12\)

\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\)

3 tháng 6 2017

Vừa vừa phải phải thôi người ta mất công gửi lên còn chửi người ta đó điên mất lịch sự

29 tháng 9 2018

a,   \(x^3+4x^2-29x+24\)

\(=x^3-x^2+5x^2-5x-24x+24\)

\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+5x-24\right)\)

\(=\left(x-1\right)\left[x\left(x-3\right)+8\left(x-3\right)\right]\)

\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)

      \(x^3+6x^2+11x+6\)

\(=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+5x+6\right)\)

\(=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

Chúc bạn học tốt.

5 tháng 7 2018

\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)=\left(x+1\right)\left(x^2+5x+6\right)\)

\(=\left(x+1\right)\left(x^2+2x+3x+6\right)=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

=(x+1)(x+2)(x+3)

16 tháng 2 2019

1) \(x^3+6x^2+11x+6\)

\(=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+5x+6\right)\)

\(=\left(x+1\right)\left(x^2+2x+3x+6\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

17 tháng 2 2019

2) \(A=n^3\left(n^2-7\right)^2-36n\)

\(A=n\left[n^2\left(n^2-7\right)^2-36\right]\)

\(A=n\left\{\left[n\left(n^2-7\right)\right]^2-6^2\right\}\)

\(A=n\left(n^3-7n-6\right)\left(n^3-7n+6\right)\)

\(A=n\left(n^3-7n-6\right)\left(n^3-n-6n+6\right)\)

\(A=n\left(n^3-7n-6\right)\left[n\left(n-1\right)\left(n+1\right)-6\left(n-1\right)\right]\)

\(A=n\left(n^3-7n-6\right)\left(n-1\right)\left(n^2+n-6\right)\)

\(A=n\left(n-1\right)\left(n^3-7n-6\right)\left(n^2+3n-2n-6\right)\)

\(A=n\left(n-1\right)\left(n^3-7n-6\right)\left[n\left(n+3\right)-2\left(n+3\right)\right]\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n^3-7n-6\right)\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n^3-n-6n-6\right)\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left[n\left(n-1\right)\left(n+1\right)-6\left(n+1\right)\right]\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left(n^2+n-6\right)\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left(n^2+3n-2n-6\right)\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left[n\left(n+3\right)-2\left(n+3\right)\right]\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left(n+3\right)\left(n-2\right)\)

\(A=\left(n-1\right)n\left(n+1\right)\left(n-2\right)^2\left(n+3\right)^2\)

Rồi sao nữa còn nghĩ :))