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17 tháng 10 2017

thế mà xl cc ak

27 tháng 8 2021

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)

4 tháng 9 2021

\(=\left(x^2+5x+8\right)\left(x^2+4x+2x+8\right)=\left(x^2+5x+8\right)\left[x\left(x+4\right)+2\left(x+4\right)\right]\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\) 

4 tháng 9 2021

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

21 tháng 10 2021

1: \(-x^2+2x+8\)

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

\(=-\left(x-4\right)\left(x+2\right)\)

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

27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

14 tháng 10 2021

a) = 2(x-2)^2

b) = 4(x - y) + (x - y)(x + y)

= (x - y)(x + y + 4)

c) = (x - 2)(x - 4)

14 tháng 10 2021

\(2\left(x-2\right)^2\)

\(\left(4+x+y\right)\left(x-y\right)\)

 

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-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

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: \(x^2+2xy+y^2+2x+2y-3\)

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

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

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

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

11 tháng 8 2021

a/ \(\left(x+y\right)^2-8\left(x+y\right)+12\)

\(=\left(x+y\right)\left(x+y-8+12\right)\)

\(=\left(x+y\right)\left(x+y+4\right)\)

==========

b/\(\left(x^2+2x\right)^2-2x^2-4x-3\)

\(=\left(x^2+2x\right)^2-\left(2x^2+4x\right)-3\)

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

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

===========

c/ \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)\left(x^2+x-2-15\right)\)

\(=\left(x^2+x\right)\left(x^2+x-17\right)\)

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16 tháng 5 2019

4 tháng 10 2021

b) \(16x-5x^2-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)

c) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

d) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)