(x-3)^3-(x+3)^3=x³-27

(1/2a+b)^3+(1/2a-b)^3=1/8a³-b³

mk chỉ làm đc 2 câu này thui

a: Ta có: \(x^2-4-\left(x+2\right)^2\)

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

=-4x-8

b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

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

=2x-1

c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)

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

\(=-3x+6\)

d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1-6x+1\right)^2\)

=4

e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)

\(=29a^2-45a-3-36a^2+24a-4\)

\(=-7a^2-21a-7\)

g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)

\(=25y^2-9-25y^2+40y-16\)

=40y-25

h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)

\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)

\(=35x^3+15x^2+15x\)

i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=16x^2\)

a: \(=4x^2+20x+25+4x^2-20x+25-\left(4x^2-1\right)\)

\(=8x^2+50-4x^2+1=4x^2+51\)

b: \(=8a^3+12a^2b+6ab^2+b^3+8a^3-12a^2b+6ab^2-b^3-16a^3\)

\(=12ab^2\)

c: \(\left(2x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)-7x^3-2x\)

\(=\left(2x-1\right)^3-x^3+8-7x^3-2x\)

\(=8x^3-12x^2+6x-1-8x^3-2x+8\)

\(=-12x^2+4x+7\)

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

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

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

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

1

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

C = 4a² - 4 - 3a - 4a² + 3a + 1

C = -3 ko phụ thuộc của x

2. ( x + 3 ) ( x - 1 ) - x ( x - 5 ) = 11

( x² + 3x - x - 3 ) - x² + 5x = 11

7x = 14

x = 2

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

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

\(\Leftrightarrow C=4a^2-4a+6a-6-3a-4a^2+3a+1\)

\(\Leftrightarrow C=\left(4a^2-4a^2\right)+\left(3a-3a\right)+\left(6a-4a\right)+\left(1-6\right)\)

\(\Leftrightarrow C=0+0+2a-5\)

\(\Leftrightarrow C=2a-5\)

Vậy giá trị của C phụ thuộc vào giá trị của biến

`Answer:`

1) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=[x\left(x+3\right)][\left(x+1\right)\left(x+2\right)]+1\)

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

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

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

2) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=[\left(12x^2+11x+0,5\right)+1,5][\left(12x^2+11x+0,5\right)-1,5]-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\)

\(=\left(12x^2+11x+0,5-2,5\right)\left(12x^2+11x+0,5+2,5\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

3) \(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)

\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)

\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)

\(=[\left(x+1\right)\left(x+7\right)][\left(x+5\right)\left(x+3\right)]+15\)

\(=\left(x^2+x+7x+7\right)\left(x^2+3x+5x+15\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(v=x^2+=8x+11\)

Đa thức có dạng sau: \(\left(v-4\right)\left(v+4\right)+15\)

\(=v^2-4^2+15\)

\(=v^2-1\)

\(=\left(v+1\right)\left(v-1\right)\)

\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

4) \(\left(x^2-a\right)^2-6x^2+4x+2a\)

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

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

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

\(=x^4-ax^2-\left(ax^2-aa\right)-6x^2+4x+2a\)

\(=x^4-2ax^2+a^2-6x^2+2a+4x\)

6) \(a^2-b^2-c^2+2bc-2a+1\)

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

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

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

7) \(4a^2-4b^2+16bc-16c^2\)

\(=4a^2-\left(4b^2-16bc+16c^2\right)\)

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

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

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

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