A= (6x-2)^2 + (2-5x)^2+2(6x-2)(2-5x)

= (6x-2)^2 +2(6x-2)(2-5x)+ (2-5x)^2

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

B= (2a^2+2a+1)(2a^2-2a+1)-(2a^2+1)^2

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

C=(x+3)(x^2-3x+9)-(54+x^3)

\(=\left(x^3+27\right)-54-x^3=27\)

D=(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)

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

E=(a+b)^2-(a-b)^2

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

Secret Personv: thật.CTV lạ z

\(C=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3-27-54-x^3=-81\)

\(C=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3+27-54-x^3=-27\)

a: \(=8x^3+y^3-8x^3+y^3=2y^3\)

b: \(=25-x^4\)

c: \(=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)

d: \(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^2\)

\(=6a^2b+2b^3-2b^2\)

e: \(=\left(x-1\right)^3\)

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

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

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

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

Bài 2: 

a) Ta có: \(A=\left(7x+5\right)^2+\left(3x-5\right)^2-\left(10-6x\right)\left(5+7x\right)\)

\(=\left(7x+5\right)^2+2\cdot\left(7x+5\right)\cdot\left(3x-5\right)+\left(3x-5\right)^2\)

\(=\left(7x+5+3x-5\right)^2\)

\(=\left(10x\right)^2=100x^2\)

Thay x=-2 vào A, ta được:

\(A=100\cdot\left(-2\right)^2=100\cdot4=400\)

b) Ta có: \(B=\left(2x+y\right)\left(y^2-2xy+4x^2\right)-8x\left(x-1\right)\left(x+1\right)\)

\(=8x^3+y^3-8x\left(x^2-1\right)\)

\(=8x^3+y^3-8x^3+8x\)

\(=8x+y^3\)

Thay x=-2 và y=3 vào B, ta được:

\(B=-2\cdot8+3^3=-16+27=11\)

Ai help mk vs

Bài 1: 

Ta có: \(A=x\left(4-x\right)\)

\(=4x-x^2\)

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

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

\(=-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x=2

Vậy: \(A_{max}=4\) khi x=2

\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)

\(=-5x-27\)

\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-\left(8x^3-y^3\right)\)

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

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

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

\(=z^2\)

a)

=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\) 

=-5x-19

b)

=\(8x^3+y^3-8x^3+y^3\) 

=\(2y^3\) 

c)

=(x+y+z-x-y)\(^2\) +x+y

=\(z^2+x+y\) 

hc tốt

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

`#3107`

`a)`

`(6x - 2)^2 + 4(3x - 1)(2 + y) + (y + 2)^2 - (6x + y)^2`

`= [(6x - 2)^2 - (6x + y)^2] + 4(3x - 1)(2 + y) + (2 + y)^2`

`= (6x - 2 - 6x - y)(6x -2 + 6x + y) + (2 + y)*[ 4(3x - 1) + 2 + y]`

`= (2 - y)(12x + y - 2) + (2 + y)*(12x - 4 + 2 + y)`

`= (2 - y)(12x + y - 2) + (2 + y)*(12x + y - 2)`

`= (12x + y - 2)(2 - y + 2 + y)`

`= (12x + y - 2)*4`

`= 48x + 4y - 8`

`b)`

\(5(2x-1)^2+2(x-1)(x+3)-2(5-2x)^2-2x(7x+12)\)

`= 5(4x^2 - 4x + 1) + 2(x^2 + 2x - 3) - 2(25 - 20x + 4x^2) - 14x^2 - 24x`

`= 20x^2 - 20x + 5 + 2x^2 + 4x - 6 - 50 + 40x - 8x^2 - 14x^2 - 24x`

`= - 51`

`c)`

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

`= [ 2(5x - 1) + x^2 - 5x + 1] * (x^2 - 5x + 1) + (5x - 1)^2 - [ (x^2)^2 - 1]`

`= (10x - 2 + x^2 - 5x + 1) * (x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`

`= (x^2 + 5x - 1)(x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`

`= x^4 - (5x - 1)^2 + (5x - 1)^2 - x^4 + 1`

`= 1`

`d)`

\((x^2+4)^2-(x^2+4)(x^2-4)(x^2+16)-8(x-4)(x+4)\)

`= (x^2 + 4)*[x^2 + 4 - (x^2 - 4)(x^2 + 16)] - 8(x^2 - 16)`

`= (x^2 + 4)(x^4 + 12x^2 - 64) - 8x^2 + 128`

`= x^6 + 16x^4 - 16x^2 - 256 - 8x^2 + 128`

`= x^6 + 16x^4 - 24x^2 - 128`