\(\left(X+1\right)\left(1+X-X^2+X^3-X^4\right)-\left(X-1\right)\lef...">
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10 tháng 8 2016

2. CM đẳng thức

a) \(a^2+b^2=\left(a+b\right)^2-2ab\)

Ta có: \(VP=\left(a+b\right)^2-2ab=a^2+2ab+b^2-2ab=a^2+b^2=VT\)

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

Ta có: \(VP=\left(a^2+b^2\right)^2-2a^2b^2=a^4+2a^2b^2+b^4-2a^2b^2=a^4+b^4=VT\)

10 tháng 8 2016

giúp mik bài 1 vs nhé

 

 

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

17 tháng 6 2019

\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)

\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)

\(=2a^2.2b^2-4a^2b^2=0\)

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

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

\(=\left[4-11x\right]^2\)

\(=16-88x+121x^2\)

chúc bn học tốt

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

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

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

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

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

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