a) \(=6a-3+15-5a=a+12\)

b) \(=25x-12x+4+35-14x=-x+39\)

d) \(=2ab+8a^2-b^2-4ab+2ab-6a^2=2a^2-b^2\)

e) \(=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4=-x^5+2x+1\)

f) \(=6y^3-3y^2+y-y+y^2-y^3-y^2+y=5y^3-3y^2+y\)

a) 3( 2a -1) +5( 3-a)

   = 3. 2a -3.1 +5. 3- 5.a

   = 6a -3+ 15-5a

   =(6a -5a )+ (-3+ 15)

b) 25x - 4(3x - 1) +7(5 - 2x)

   = 25x -4.3x + 4.1 + 7.5 - 7.2

   =25x - 12x + 4 +35 - 14x

   = (25x-12x-14x)+(4+35)

   = -x=39

c) -12x3 -x1-2x-18x2

   = -36x-x-2x-36x

   = -75x

d) (2a-b)(b+4a)+2a(b-3a)

   = 2ab+2a4a-bb-b4a+2ab-2a3b

   = 2ab+8a²-b²-4ab+2ab-6a²

   =(2ab-4ab+2ab)+(8a²-6a²)-b²

   = 2a²-b²

e) (x+1)(2+x-x2+x3-x4)

   = (x+1)(2-2x)

   = x2-x2x+1.2-1.2x

   =(2x-2x)-2x²+2

   = -2x²+2

 (a-b)^3+(b-a)^3

=(a-b+b--a)[(a-b)^2-(a-b)(b-a)+(b-a)^2]

=0

1) Ta có: \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)+ \(3xyz\)

Mà x+y+z=0

=> \(x^3+y^3+z^3=3xyz\)

( ko thể = 3xy²)

2)  Ta có: \(A=\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)+1\) 

                    = \(\left(n+1\right)\left(n+4\right)\cdot\left(n+2\right)\left(n+3\right)+1\)

                     = \(\left(n^2+5n+4\right)\left(n^2+5n+6\right)+1\)

Đặt t= \(n^2+5n+5\)

=> A= \(\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\) là 1 số chính phương.

đề là x²+x+1=0 đúng ko

<=>x²+2.1/2x+1/4+3/4=0

<=>(x+1/2)²+3/4=0( vô lí )

vậy ko có gt của x

a)0,75

b)-2,274541137

c)-0,263157894