\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)

Nguyễn Đình Dũng vs Minh Hiền không khác nhau tẹo nào 

a. ( 6x + 1 ) ² + (6x - 1 )² - 2(1+ 6x ) (6x-1 )

= (6x+1)²-2(6x+1)(6x-1)+(6x-1)²

=[(6x+1)-(6x-1)]²

= (6x+1-6x+1)²

= 2²

=4

=[ (6x+1)-(6x-1) ]²

1.ta có: 
x^3 + y^3 + z^3 - 3xyz = (x+y)^3 + z^3 - 3x^2y - 3xy^2 - 3xyz 
= (x+y)^3 + z^3 - 3xy(x + y + z) 
= (x+y+z)^3 - 3(x+y)^2.z - 3(x+y)z^2 - 3xy(x + y + z) 
= (x+y+z)^3 - 3(x+y)z(x+ y + z) - 3xy(x + y + z) 
=(x+y+z)[(x+y+z)^2 - 3(x+y)z - 3xy] 
với x+y+z = 0 => x^3 + y^3 + z^3 - 3xyz = 0 => x^3 + y^3 + z^3 = 3xyz

2.

x=5

=>6=x+1

=> A=x⁶-6x⁵+6x⁴-6x³+6x²-6x+6=x⁶-(x+1).x⁵+(x+1)x⁴-(x+1)x³+(x+1)x²-(x+1)x+(x+1)

=x⁶-x⁶-x⁵+x⁵-x⁴+x⁴-x³+x³-x²+x²-x+x+1

=1

vậy A=1 khi x=5

1,

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=3abc\)

2,

\(A=\left(x-1\right)\left(x-5\right)\left(x^4+x^2+1\right)+1\)

x=5 thì A=1

a+b+c=0 =>a+b= -c

=>  (a+b)^3=-c^3

=>a^3+b^3+c^3= -3ab(a+b)

=>a^3+b^3+c^3=3abc      (vì a+b=c)

=> dpcm

a)  Ta có:(6x+1)^2 +(6x-1)^2 +2(6x+1)(6x-1) =[(6x+1)+(6x-1)]^2 =(12x)^2=(12^2)(x^2)=144.x^2

b)  Ta có:(6x+1)^2 +(6x-1)^2 -(12x+2)(6x-1)=(6x+1)^2 +(6x-1)^2 -2(6x+1)(6x-1)=[(6x+1)-(6x-1)]^2=2^2=4

c)  Ta có:(ac+bd)^2 +(ad-bc)^2=(ac)^2 +2(ac)(bd) +(bd^2) +(ad)^2 -2(ad)(bc) +(bc)^2

=a^2.c^2 +2abcd +b^2 d^2 +a^2.d^2 -2abcd +b^2.c^2=a^2.c^2 +b^2.d^2 +a^2.d^2 +b^2.c^2

=(a^2 +b^2)(c^2 +d^2)

d)  Ta có:(ac-bd)(ac+bd)=(ac)^2 -(bd)^2=a^2.c^2 -b^2.d^2

Rút gọn nha các cậu

\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\times\frac{x^2-36}{12x^2+12}\)

\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\times\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)

\(A=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}\times\frac{1}{12\left(x^2+1\right)}\)

\(A=\frac{12\left(x^2+1\right)}{x}\times\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

a) \(6x^3-6x=0\Leftrightarrow6x\left(x^2-1\right)=0\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)b) \(2x\left(3x+7\right)-6x^2=28\Leftrightarrow6x^2+14x-6x^2=28\Leftrightarrow14x=28\Leftrightarrow x=2\)

c) \(2\left(4x+4\right)-5\left(x-3\right)=0\Leftrightarrow8x+8-5x+15=0\Leftrightarrow3x=-23\Leftrightarrow x=-\dfrac{23}{3}\)

a: Ta có: \(6x^3-6x=0\)

\(\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

b: Ta có: \(2x\left(3x+7\right)-6x^2=28\)

\(\Leftrightarrow6x^2+14x-6x^2=28\)

\(\Leftrightarrow14x=28\)

hay x=2

c: Ta có: \(2\left(4x+4\right)-5\left(x-3\right)=0\)

\(\Leftrightarrow8x+8-5x+15=0\)

\(\Leftrightarrow3x=-23\)

hay \(x=-\dfrac{23}{3}\)