có

TL

z²

HT

TL:

z²

~HT~

\(\left(x-3\right)^3+\left(3-x\right)\)

\(=\left(x-3\right)\left[\left(x-3\right)^2-1\right]\)

\(=\left(x-3\right)\left(x-2\right)\left(x-4\right)\)

\(1)\)\(4x^2-2x=0\)

\(2x\left(2x-1\right)=0\)

\(\orbr{\begin{cases}2x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\div2\\2x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0,5\end{cases}}}\)

\(2)\)\(\left(x-5\right)^2-9=0\)

\(\left(x-5\right)^2=0+9\)

\(\left(x-5\right)^2=9\)

\(\left(x-5\right)^2=3^2\)

\(x-5=3\)

\(x=3+5\)

\(x=8\)

\(3)\)\(5x\left(x-2\right)-\left(2-x\right)=0\)

\(5x\left(x-2\right)+\left(x-2\right)=0\)

\(\left(x-2\right)\left(5x+1\right)=0\)

\(\orbr{\begin{cases}x-2=0\\5x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0+2\\5x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-\frac{1}{5}\end{cases}}}\)

\(x^2+5x-2\)

Vì đa thức có 2 nghiệm nên có thể phân tích thành

\(\left(x+\frac{5-\sqrt{33}}{2}\right)\left(x+\frac{5+\sqrt{33}}{2}\right)\)

\(\left(x-3\right)^3-\left(x+2\right)\left(x^2-2x+4\right)-\left(3-2x\right)\left(3x-2\right)\)

\(=\left(x^3-9x^2+27x-27\right)-\left(x^3+8\right)-\left(9x-6x^2-6+4x\right)\)

\(=-x-10\)

\(\left(3x-1\right)\left(x+2\right)-\left(2-x\right)\left(x+3\right)\)

\(=\left(3x^2-x+6x-2\right)-\left(2x-x^2+6-3x\right)\)

\(=\left(3x^2+x^2\right)+\left(5x+x\right)-2-6=4x^2+6x-8=-10\)

\(50^2-49^2+48^2-47^2+...+2^2-1^2\)

\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)

\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=50+49+...+2+1\)

\(=\frac{50\left(50+1\right)}{2}=1275\)