xin lỗi nhưng thật sự đề sai vãi ca chưởng rồi bạn ơi

đề đúng nha bn mk chỉ cần giải câu c thôi nha chứ đề k hề sai

ai làm đúng mị cho 2 k

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

\(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

\(x^6-y^6=\left(x-y\right)\left(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5\right)\)

Tham khảo nhé~

\(\frac{2}{3-x}+\frac{4x}{x^2-9}\)

\(=\frac{2}{3-x}-\frac{4x}{9-x^2}\)

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

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

\(=\frac{2x+6-4x}{\left(3-x\right)\left(3+x\right)}\)

\(=\frac{6-2x}{\left(3-x\right)\left(3+x\right)}\)

\(=\frac{2.\left(3-x\right)}{\left(3-x\right)\left(3+x\right)}\)

\(=\frac{2}{x+3}\)

\(\frac{2}{3-x}+\frac{4x}{x^2-9}\)

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

\(ĐKXĐ:\hept{\begin{cases}x-3#0\\x+3#0\end{cases}}\Rightarrow\hept{\begin{cases}x#3\\x#-3\end{cases}}\)

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

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

\(=\frac{-2x-6+4x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x\left(-1+2\right)-6}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x-6}{\left(x-3\right)\left(x+3\right)}\)

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

\(=\frac{2}{x+3}\)

Vì sợ bạn ko hiểu nên mình mới làm dài dòng thế ok

\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)

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

\(x^2-3x+xy-3y\)

\(=x\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(x-3\right)\)

\(x^2-2xy+y^2-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)

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

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

\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

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

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

Tham khảo nhé~

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

\(\frac{125x^3+1}{5x+1}=\frac{\left(5x\right)^3+1}{5x+1}=\frac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)

\(\frac{2x^3+5x^2-2x+3}{2x^2-x+1}=\frac{\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)}{2x^2-x+1}\)

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

Tham khảo nhé~

\(\left(3x-4\right)^2-36=0\)

\(\left(3x-4\right)^2-6^2=0\)

\(\left(3x-4-6\right)\left(3x-4+6\right)=0\)

\(\left(3x-10\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-10=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=\frac{10}{3}\\x=-\frac{2}{3}\end{cases}}\)