Ta có: \(x+\frac{1}{x}=3\Leftrightarrow\left(x+\frac{1}{x}\right)^2=3^2\Leftrightarrow x^2+2+\frac{1}{x^2}=9\Leftrightarrow x^2+\frac{1}{x^2}=7\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)^2=7^2\Leftrightarrow x^4+2+\frac{1}{x^4}=49\Leftrightarrow x^4+\frac{1}{x^4}=47\)

\(x+\frac{1}{x}=3\)

=>  \(\left(x+\frac{1}{x}\right)^2=9\)

<=> \(x^2+\frac{1}{x^2}+2=9\)

<=> \(x^2+\frac{1}{x^2}=7\)

=>   \(\left(x^2+\frac{1}{x^2}\right)^2=49\)

<=> \(x^4+\frac{1}{x^4}+2=49\)

<=>  \(x^4+\frac{1}{x^4}=47\)

gọi vận tốc là x(km/h), thời gian là: y (giờ) đk(x,y>0)

quãng đường AB là:xy(km)

theo đề ta có hệ phương trình:

_{^{\(\hept{\begin{cases}\left(x+10\right).\left(y-2\right)=xy\\\left(x-10\right).\left(y+3\right)=xy\end{cases}}\)}}=>\(\hept{\begin{cases}xy-2x+10y-20=xy\\xy+3x-10y-30=xy\end{cases}}\)=>\(\hept{\begin{cases}-2x+10y=20\\3x-10y=30\end{cases}}\)=>\(\hept{\begin{cases}x=50\\y=12\end{cases}}\)

=> quãng đường AB là :50.12=600(km)

http://123link.pro/TuLVQ15F

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

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

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

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

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

\(=\left(x-1\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)\)

phân tích đa thức thành nhân tử

\(-50x^2y^2+2\left(x-y\right)^2\)

\(=2\left[\left(x-y\right)^2-25x^2y^2\right]\)

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

p/s: chúc bạn học tốt

\(x^4+x^2y^2+y^4\)

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

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

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