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a ) \(\left(x+2\right)^3-\left(x-2\right)^3\)
\(=\left[\left(x+2\right)-\left(x-2\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right]\)
\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)
\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)
\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)
\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)
\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)
\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)
\(\Leftrightarrow4x^2+6x-51=0\)
\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)
\(\left(x^2-4x\right)^2+2\left(x-2\right)^2=43\)
\(\Leftrightarrow x^4-8x^3+16x^2+2x^2-8x+8-43=0\)
\(\Leftrightarrow x^4-8x^3+18x^2-8x-35=0\)
\(\Leftrightarrow x^4+x^3-9x^3-9x^2+27x^2+27x-35x-35=0\)
\(\Leftrightarrow x^3\left(x+1\right)-9x^2\left(x+1\right)+27x\left(x+1\right)-35\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-9x^2+27x-35\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-5x^2-4x^2+20x+7x-35\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x-5\right)-4x\left(x-5\right)+7\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-5\right)\left(x^2-4x+7\right)=0\)
Vì \(x^2-4x+7< 0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}}\)
Vậy....