b) \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)

\(\Leftrightarrow x^3+3^3-x^3-2x=0\)

\(\Leftrightarrow27-2x=0\)

\(\Leftrightarrow2x=27\)

\(\Leftrightarrow x=\frac{27}{2}=13,5\)

c) \(\left(x-2\right)^3-x^3+6x^2=7\)

\\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3-x^3+6x^2=7\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-7=0\)

\(\Leftrightarrow12x=15\Leftrightarrow x=\frac{15}{12}=\frac{5}{4}=1,25\)

\(\left(x+2\right)^2-\left(x-3\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-\left(x^2-6x+9\right)=7\)

\(\Leftrightarrow x^2+4x+4-x^2+6x-9=7\)

\(\Leftrightarrow10x=7-4+9\)

\(\Leftrightarrow10x=12\)

\(\Leftrightarrow x=\frac{12}{10}=\frac{6}{5}\)

\(x^4+4x^3-16x-16=0\)

\(\Leftrightarrow\left(x^4-16\right)+\left(4x^3-16x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)+4x\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+4+4x\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)^3=0\)

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\\left(x+2\right)^3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-2\end{cases}}}\)

=.= hok tốt!!

x^4-45*x^3+112*x^2+288*x+256

tk nhé

wtf cái này toán lớp 7 mà
ta có LK là trung tuyến ( do K là trọng tâm ) mà trong tam giác cân thì trung tuyến trùng đường cao >> LK là đường cao >> vuông góc AC