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

b. \(\left(a+2\right)\left(a-2\right)\left(a^2+2a+4\right)\left(a^2-2x+4\right)=a^6-64\)

c. \(\left(2+3y\right)^2-\left(2x-3y\right)^2-12xy=4+12y-4x^2\)

d. \(\left(x+1\right)^3-\left(x-1\right)^3-\left(x^3-1\right)-\left(x-1\right)\left(x^2+x+1\right)=-2x^3+6x^2+4\)

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

\(B=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)

\(C=9y^2+12y+4-\left(4x^2-12xy+9y^2\right)-12xy=12y+4-4x^2\)

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

giải:

a

[a^2+ (a-1)]*[(a^2-(a-1)]=(a^2)^2-(a-1)^2=a^2-(a-1)^2
 

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

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a: \(=x^2+10xy+25y^2-5\left(x^3-9x^2y+27xy^2-27y^3\right)\)

\(=x^2+10xy+25y^2-5x^3+45x^2y-135xy^2+135y^3\)

b: \(=9x^2-24x+16-2x+22\)

\(=9x^2-26x+38\)

c: \(=x^3+8-x^2+2=x^3-x^2+10\)

d: \(=\left(a^3-8\right)\left(a^3+8\right)=a^6-64\)

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

b)\(=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)

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

d)\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6=-8\)

e) \(=4+12y+9y^2-4x^2+12xy-9y^2-12xy=-4x^2+12y+4\)

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

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

cảm ơn nhé

Rút gọn biểu thức sau : a) (a2 +a- 1) ( a2 - a+1)

b) ( a+2) (a -2) ( a2+ 2a +4) ( a2- 2a +4)

c) (x-2)2 -(x - 3) (x-1)

d) (x-1)3 - (x+1)3 + 6(x+1) (x -1)

e) (2 +3y)2- ( 2x- 3y)2 - 12xy

g) (x+ 1)3 - (x-1 )3- (x3 -1 )- (x- 1)(x2 +x +1)

h) (x+1) (x2 -x+ 1)-( x-1) (x2 + x +1 )

Giúp mình nhé ><

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

chịu

a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)

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

d: \(=9x^2+6x+1-9x^2+6x-1=12x\)

a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)

e: \(=x^3+1-x^3+1=2\)