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\(a,\left(2x-3\right)^3-\left(x-1\right)^3-2\left(x-2\right)\left(x+2\right)\)
\(=8x^3+36x^2+27x+27-\left(x^3-3x^2+3x-1\right)-2\left(x^2-4\right)\)
\(=8x^3+36x^2+27x+27-x^3+3x^2-3x+1-2x^2+8\)
\(=7x^3+37x^2+24x+36\)
\(b,\left(x-3\right)^2-2\left(x+2\right)^3-4\left(x+3\right)\left(x-3\right)\)
\(=x^2-6x+9-2\left(x^3+6x^2+12x+8\right)-4\left(x^2-9\right)\)
\(=x^2-6x+9-2x^3-12x^2-24x-16-4x^2+36\)
\(=-15x^2-30x-2x^3+45\)
\(c,\left(2x-5\right)^3-4\left(x-2\right)\left(x+2\right)-2\left(x+1\right)^3\)
\(=8x^3-10x^2+50x-25-4\left(x^2-4\right)-2\left(x^3+3x^2+3x+1\right)\)
\(=8x^3-10x^2+50x-25-4x^2+16-2x^3-6x^2-6x-2\)
\(=6x^3-20x^2+44x-11\)
(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
Bài 1.
1) ( x - 1 )3 - x( x - 3 )2 + 1
= x3 - 3x2 + 3x - 1 - x( x2 - 6x + 9 ) + 1
= x3 - 3x2 + 3x - x3 + 6x2 - 9x
= 3x2 - 6x = 3x( x - 2 )
2) ( x + 2 )2 - x2( x + 6 )
= x2 + 4x + 4 - x3 - 6x2
= -x3 - 5x2 + 4x + 4
3) ( x + 2 )3 - ( x - 2 )3
= x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 )
= x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8
= 12x2 + 16 ( có phụ thuộc vào biến )
Bài 2.
1) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
2) ( x - 2 )3 - x( x + 1 )( x - 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
Bài 1:
a) \(\left(x-1\right)^3-x\left(x-3\right)^2+1\)
\(=x^3-3x^2+3x-1-x^3+6x^2-9x+1\)
\(=3x^2-6x\)
b) \(\left(x+2\right)^2-x^2\left(x+6\right)\)
\(=x^2+4x+4-x^3-6x^2\)
\(=-x^3-5x^2+4x+4\)
c) \(\left(x+2\right)^3-\left(x-2\right)^3\)
\(=x^3+6x^2+12x+8-x^3+6x^2-12x+8\)
\(=12x^2+16\)
=> BT phụ thuộc vào biến
\(\left(x+3\right)^2+\left(x-3\right)^2-2.\left(x+3\right).\left(x-3\right)\)
\(=\left(x^2+6x+9\right)+\left(x^2-6x+9\right)-2.\left(x^2-9\right)\)
\(=x^2+6x+9+x^2+x^2-6x+9-2x^2+18\)
\(=36\)