2: Đặt \(A=\left(x-3\right)^3-\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=x^3-9x^2+27x-27-8x^3-1\)
\(=-7x^3-9x^2+27x-28\)
Đặt \(B=\left(x+2\right)^3-\left(2x-3\right)^3-18x\left(2x-3\right)\)
\(=x^3+6x^2+12x+8-8x^3+36x^2-54x+27-36x^2+54x\)
\(=-7x^3+6x^2+12x+35\)
=>\(-9x^2+27x-28=6x^2+12x+35\)
=>\(-15x^2+15x-63=0\)
=>\(x\in\varnothing\)
3: \(\Leftrightarrow x^3+6x^2+12x+8-8x^3-36x^2-54x-27+4x^2-9=x^3-8+6x^2+12x\)
\(\Leftrightarrow-7x^3-26x^2-42x-28=x^3+6x^2-8+12x\)
\(\Leftrightarrow-8x^3-32x^2-54x-20=0\)
=>x=-1/2
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