d) \(x\left(x^2+3x-2\right)=x^3+100+3x^2\)

\(\Rightarrow x^3+3x^2-2x=x^3+100+3x^2\\ \Rightarrow-2x=100\\ \Rightarrow x=-50\)

e) \(6x\left(x-1\right)=\left(2x+1\right)\left(3x-2\right)\)

\(\Rightarrow6x^2-6x=6x^2-x-2\\ \Rightarrow5x=2\\ \Rightarrow x=\dfrac{2}{5}\)

f) \(\left(x+1\right)^3-\left(x^3+3x^2+4\right)=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2-4=0\\ \Rightarrow3x=3\\ \Rightarrow x=1\)

a) \(\left(x+4\right)^2-x\left(x+10\right)=0\)

\(\Rightarrow x^2+8x+16-x^2-10x=0\\ \Rightarrow-2x=-16\\ \Rightarrow x=8\)

b) \(\left(x-5\right)^2-\left(x+4\right)\left(x-6\right)=0\)

\(\Rightarrow x^2-10x+25-x^2+6x-4x+24=0\\ \Rightarrow-8x=-49\\ \Rightarrow x=\dfrac{49}{8}\)

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

\(\Rightarrow4x^2-8x=4x^2-36\\ \Rightarrow-8x=-36\\ \Rightarrow x=\dfrac{9}{2}\)

\(\left(2x^3-3x-1\right)\left(5x+2\right)\\ =10x^4-15x^2-5x+4x^3-6x-2\\ =10x^4+4x^3-15x^2-11x-2\)

a) Áp dụng định lý Pytago ta có: 

\(BC^2=AB^2+AC^2=5^2+12^2=169\Rightarrow BC=13\left(cm\right)\)

b) Ta có: \(\sin B=\dfrac{AC}{BC}=\dfrac{5}{13}\Rightarrow\widehat{B}\approx22,6^0\Rightarrow\widehat{C}\approx67,4^0\)

Tổng của hai số là: \(72\dfrac{2}{3}=\dfrac{218}{3}\)

Số thứ nhất là: \(\dfrac{218}{3}:\left(2+5\right)\times2=\dfrac{436}{21}\)

Số thứ hai là: \(\dfrac{218}{3}-\dfrac{436}{21}=\dfrac{1090}{21}\)

\(3\dfrac{2}{5}=\dfrac{3\times5+2}{5}=\dfrac{17}{5}\)

\(7\dfrac{1}{8}=\dfrac{7\times8+1}{8}=\dfrac{57}{8}\)

\(4\dfrac{5}{7}=\dfrac{4\times7+5}{7}=\dfrac{33}{7}\)

\(4357\times26+\left(6378-10208:58\right)\\ =113282+6378-176\\ =119660-176\\ =119484\)