a) \(\dfrac{x^2-1}{120}+\dfrac{x^2-2}{119}+\dfrac{x^2-3}{118}=3\)
\(=\dfrac{x^2-1}{120}-1+\dfrac{x^2-2}{119}-1+\dfrac{x^2-3}{118}-1=0\)\(=\dfrac{x^2-121}{120}+\dfrac{x^2-121}{119}+\dfrac{x^2-121}{118}=0\)
\(=\left(x^2-121\right).\left(\dfrac{1}{120}+\dfrac{1}{119}+\dfrac{1}{118}\right)=0\)
\(=\left(x+11\right)\left(x-11\right)\left(\dfrac{1}{120}+\dfrac{1}{119}+\dfrac{1}{118}\right)=0\)
⇒\(\left[{}\begin{matrix}x+11=0\\x-11=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)
b)\(\)
\(x^3-5x^2+8x-4\)
\(=x^3-6x^2+12x-8+x^2-4x+4\)
\(=\left(x-2\right)^3+\left(x-2\right)^2\)
\(=\left(x-2\right)^2\left(x-2+1\right)=\left(x-2\right)^2\left(x-1\right)\)
c)
\(=\left(x-1\right)^3+\left(x-2\right)^3+\left(3-2x\right)^3=0\)
\(=\left[\left(x-1\right)+\left(x-2\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(x-2\right)+\left(x-2\right)^2\right]+\left(3-2x\right)^3=0\)
\(=\left(2x-3\right)\left[x^2-2x+1-\left(x^2-3x+2\right)+x^2-4x+4\right]+\left(3-2x\right)^3=0\)
\(=\left(2x-3\right)\left(x^2-3x+3\right)-\left(2x-3\right)^3=0\)
\(=\left(2x-3\right)\left[\left(x^2-3x+3\right)-\left(2x-3\right)^2\right]=0\)
\(=\left(2x-3\right)\left(-3x^2+9x-6\right)=0\)
\(=-3\left(x^2-3x+2\right)\left(2x-3\right)=0\)
\(=-3\left(x-1\right)\left(x-2\right)\left(2x-3\right)=0\)
⇒\(\left[{}\begin{matrix}x-1=0\\x-2=0\\2x-3=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
