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

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

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

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

\(b,4x^3+9x^2-19x-30=3x^3+x^3+9x^2-9x-10x-30\)

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

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

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

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

2)\(\left(2x-3\right)^2-\left(x+5\right)^2=0\)\(\Leftrightarrow\left[\left(2x-3\right)-\left(x+5\right)\right]\left[\left(2x-3\right)+\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0+8=8\\x=\frac{0-2}{3}=\frac{-2}{3}\end{cases}}\)

bạn viết lại đề bài cho rõ đc ko b 

bạn ko hieu cho nao

\(3n^2-13n+29=3n.\left(n-3\right)-4n+29\)

\(=3n.\left(n-3\right)-4.\left(n-3\right)+17=\left(3n-4\right).\left(n-3\right)+17\)

=> đề \(3n^2-13n+29⋮n-3\Rightarrow17⋮n-3\Rightarrow n-3\inƯ\left(17\right)=\left\{\pm1,\pm17\right\}\)

=> \(n\in\left\{4,2,-14,20\right\}\)

vì n là số nguyên dương => n\(\in\){4,2,20}

Câu này ns chung đễ thui 

XD

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

\(\Leftrightarrow x\left(x-10\right)+1\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-1\end{cases}}\)

\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)=x+y+z\)

<=>\(\frac{x^2+x\left(y+z\right)}{y+z}+\frac{y^2+y\left(z+x\right)}{z+x}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)

<=>\(\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)

<=>\(S=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)

 x/(y+z)+y/(x+z)+z/(x+y)=1

=>\(\frac{x^2}{\left(y+z\right)^2}\)+\(\frac{y^2}{\left(x+z\right)^2}\)+\(\frac{z^2}{\left(x+y\right)^2}\)+2(\(\frac{xy}{\left(y+z\right)\cdot\left(x+z\right)}\)+\(\frac{yz}{\left(x+z\right)\left(x+y\right)}\)+\(\frac{zx}{\left(z+y\right)\cdot\left(x+y\right)}\))=1