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1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)
\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)
\(=-63x^2-21x-42\)
2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)
3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)
1. \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3+125-\left(x^3-8\right)=x^3+125-x^3+8=133\)
1,
\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\\ =\left(x^3+5^3\right)-\left(x^3-2^3\right)\\ =x^3+125-x^3+8\\ =\left(x^3-x^3\right)+\left(125+8\right)\\ =133\)
b,
\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+1\right)^3\\ =\left[\left(2x\right)^3-3^3\right]-\left[\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x+1+1\right]\\ =\left(8x^3-27\right)-\left(8x^3+12x^2+6x+1\right)\\ =8x^3-27-8x^3-12x^2-6x-1\\ =\left(8x^3-8x^3\right)-\left(12x^2+6x\right)-\left(27+1\right)\\ =-6x\left(2x+1\right)-28\\ =\left(-2\right)\left[3x\left(2x+1\right)+14\right]\)
1. (3x - 5)2 - (3x + 1)2 = 8
=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8
=> -6(6x - 4) = 8
=> 6x - 4 = \(\dfrac{-4}{3}\)
\(\Rightarrow x=\dfrac{4}{9}\)
2) 2x(8x - 3) - (4x - 3)2 = 27
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> x = 2
3) (2x - 3)2 - (2x + 1)2 = 3
=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3
=> -4(4x - 2) = 3
=> 4x - 2 = \(\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{5}{16}\)
4) (x + 5)2 - x2 = 45
=> (x + 5 - x)(x + 5 + x) = 45
=> 5(2x + 5) = 45
=> 2x + 5 = 9
=> x = 2
5) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 18
=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9(x2 + 2x + 1) = 18
=> -9x2 + 27x + 9x2 + 18x + 9 = 18
=> 45x + 9 = 18
=> 45x = 9
=> x = \(\dfrac{1}{5}\)
6) x(x - 4)(x + 4) - (x - 5)(x2 + 5x + 25) = 13
=> x (x2 - 16) - (x3 - 125) = 13
=> x3 - 16x - x3 + 125 = 13
=> -16x = -112
=> x = 7.
a: \(=2x^2-x+5\)
b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c: \(=-x^3+\dfrac{3}{2}-2x\)
d: \(=-2x^2+4xy-6y^2\)
e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
công thức mở rộng câu thứ 3:\(\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right)\)