\(\left(-x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x+7=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}\)

2x³+5x²=7x

-7x+5x²+2x³=7x+(-7x)

ta có 7x+(-7x)

-7x + 5x2 + 2x3 = 0

x(-7 + 5x + 2x2) = 0

x[(-7 + -2x)(1 + -1x)] = 0

2x^3 + 5x^2 = 7x

<=> 2x^3 + 5x^2 - 7x = 0

<=> x(2x^2 + 5x - 7) = 0

<=> x(2x^2 + 7x - 2x - 7) = 0

<=> x[x(2x + 7) - (2x + 7)] = 0

<=> x(x - 1)(2x + 7) = 0

<=> x = 0 hoặc x - 1 = 0 hoặc 2x + 7 = 0

<=> x = 0 hoặc x = 1 hoặc x = -7/2

\(\frac{x^3-27}{x^2-9}=0\)

\(\Leftrightarrow x^3-27=0\)

\(\Leftrightarrow x^3=27\)

\(\Leftrightarrow x=\pm3\)

ĐKXĐ :\(x\ne\pm3\)

\(C=A+B=\frac{x^4+1}{x^4-x^3+2x^2-x+1}+\frac{x}{x^2-x+1}\)

\(=\frac{x^4+1}{x^4-x^3+x^2+x^2-x+1}+\frac{x}{x^2-x+1}\)

\(=\frac{x^4+1}{\left(x^2+1\right)\left(x^2-x+1\right)}+\frac{x^3+x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x^4+x^3+x+1}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x^3+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)

\(=\frac{\left(x+1\right)^2}{x^2+1}\)

\(C=0\Leftrightarrow\frac{\left(x+1\right)^2}{x^2+1}=0\)

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