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Ta có 2x(2x + 1)2 - 3x(x + 3)(x - 3) - 4x(x + 1)2
= 2x(4x2 + 4x + 1) - 3x(x2 - 9) - 4x(x2 + 2x + 1)
= 8x3 + 8x2 + 2x - 3x3 + 27x - 4x3 - 8x2 - 4x
= 8x3 - 3x3 - 4x3 + 8x2 - 8x2 + 2x + 27x - 4x
= x3 + 25x
a oi hinh nhu sai r con +16x2 nua co a , anh tinh lai ho e duoc kh
a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)
\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)
\(=-19x^2+16x-4\)
\(a,\)\(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2.\)
\(=\left[\left(x-y\right)+\left(x+y\right)\right]^2=\left(x-y+x+y\right)^2=x^2\)
\(b,\)\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(54+8x\right)\)
\(=8x^2-27-54-8x=8x^2-8x-81\)
\(c,\)\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=27x^3+y^3-\left(27x^3-y^3\right)=2y^3\)
\(d,\)\(\left(a+b+c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=a^2+b^2+c^2+2ab+2bc+2ac-a^2+2ac-c^2-2ab+2bc\)
\(=b^2+4bc+4ac\)
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: Để A>0 thì x-3>0
hay x>3
=\(-19\frac{1}{4}\)