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28 tháng 8 2018

a) \(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)

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

\(=\left(x+1\right)^4+x^2\left(x+1\right)^2+2x\left(x+1\right)+1\)

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

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

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

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

b) \(\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt \(x=a+b-2c\)

\(y=b+c-2a\)

\(z=c+a-2b\)

\(\Rightarrow x+y+z=a+b-2c+b+c-2a+c+a-2b\)

\(\Rightarrow x+y+z=0\)

\(\Rightarrow x+y=-z\left(1\right)\)

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

\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=\left(-z\right)^3\)

\(\Rightarrow x^3+y^3+z^3+3xy.\left(-z\right)=0\) ( Vì x + y = -z )

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Rightarrow\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3=3\left(a+b-2c\right)\left(b+c-2a\right)\left(c+a-2b\right)\)

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

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

\(=x^4-2x^3+6x^2-8x+8\)

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

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

d) \(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+10^2-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2\)

\(=\left(x^2+10-6x\right)\left(x^2+10+6x\right)\)

28 tháng 9 2017

bạn viết so mu nhu the nao day ?

28 tháng 9 2017

a)x2-2xy+y2+3x-3y-10

=(x2-2xy+y2)+(3x-3y)-10

=(x-y)2+3(x-y)-10

=(x-y).(x-y+3)-10

7 tháng 7 2019

2) Để sau đi (em chưa nghĩ ra)

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

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

Đặt x - y = a; y - z = b => z - x = -(a+b)

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

\(=a\left[\left(x+y\right)^2-\left(z+x\right)^2\right]+b\left[\left(y+z\right)^2-\left(z+x\right)^2\right]\)

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

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

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

Em tính sai sót chỗ nào thì thông cảm cho em ạ :>

3 tháng 11 2017

1) a4+b4+c42a2b22a2c22b2c2

=2(a4+b4+c4-4a2b2-4a2c2-4b2c2)

=2a4+2b4+2c4-4a2b2-4a2c2-4b2c2

=(a4-2a2b2+b4)+(a4-2a2c2+c4)+(b4-2b2c2+c4

1 tháng 10 2017

a)\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)

\(=\left(a-b\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-b^3\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)

26 tháng 5 2017

1. (a2+b2+ab)2-a2b2-b2c2-c2a2

=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2

=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2

=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)

=(a2+b2)[(a+b)2-c2]

=(a2+b2)(a+b+c)(a+b-c)

2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2

3. a(b3-c3)+b(c3-a3)+c(a3-b3)

=ab3-ac3+bc3-ba3+ca3-cb3

=a3(c-b)+b3(a-c)+c3(b-a)

=a3(c-b)-b3(c-a)+c3(b-a)

=a3(c-b)-b3(c-b+b-a)+c3(b-a)

=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)

=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)

=(a-b)(c-b)(a2+ab+2b2+bc+c2)

4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)

5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]

=2b(3a2+b2)

6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]

=(x-y-1)(x2+y2+xy-2x-y+1)

7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)

(Đúng nhớ like nhá !)

26 tháng 5 2017

Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi

22 tháng 5 2018

a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)

b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)

22 tháng 5 2018

lm tiếp câu c

c)  \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)

\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)

\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)

Đặt   \(x^2-9x+17=a\) ta có:

        \(C=\left(a-3\right)\left(a+3\right)-72\)

            \(=a^2-9-72\)

           \(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được:  \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)

          

18 tháng 6 2017

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

Đặt \(t=x^2+3x\) thì biểu thức có dạng \(t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)

b)\(\left(x^2-x+2\right)^2+4x^2-4x-4=\left(x^2-x+2\right)^2+4\left(x^2-x-1\right)\)

Đặt \(k=x^2-x+2\) thì biểu thức có dạng

k2+4(k-3)=k2+4k-12=k2-2k+6k-12=k(k-2)+6(k-2)=(k-2)(k+6)=(x2-x)(x2-x+8)=(x-1)x(x2-x+8)

c)làm tương tự câu a