\(e,-5x+x^2-14\)

\(=x^2+2x-7x-14\)

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

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

\(f,x^3+8+6x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+8x+4\right)\)

\(g,15x^2-7xy-2y^2\)

\(=15x^2+3xy-10xy-2y^2\)

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

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

\(h,3x^2-16x+5\)

\(=3x^2-x-15x+5\)

\(=x\left(3x-1\right)+5\left(3x-1\right)\)

\(=\left(3x-1\right)\left(x+5\right)\)

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

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

\(b,4x^2-9y^2+4x-6y\)

\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)

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

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

\(c,-x^2+5x+2xy-5y-y^2\)

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

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

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

\(d,x^2+4x-12\)

\(=x^2-2x+6x-12\)

\(=x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

a) \(x^2-xy+x=x\left(x-y+1\right)\)

b) \(x\left(x-y\right)-2\left(y-x\right)\) 

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

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

c) \(9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)

d) \(x^2-xy-4x+2y+4\)

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

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

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

Chuc ban hoc tot !!!

Thank bạn nhé!!!.:))

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)

\(P=\left(\frac{x^2}{x^3-4x}-\frac{10}{5x+10}-\frac{1}{2-x}\right):\)\(\left(x+2+\frac{6-x^2}{x-2}\right)\)

\(=\left(\frac{x^2}{x\left(x^2-4\right)}-\frac{10}{5\left(x+2\right)}+\frac{1}{x-2}\right)\)\(:\left(\frac{\left(x-2\right)\left(x+2\right)}{x-2}+\frac{6-x^2}{x-2}\right)\)

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

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

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

\(b,P\in Z\Leftrightarrow\frac{3}{x+2}\in Z\Rightarrow3\)\(⋮\)\(x+2\Rightarrow x+2\inƯ_3\)

MÀ \(Ư_3=\left\{\pm1;\pm3\right\}\)

TH1 : \(x+2=-1\Rightarrow x=-3\)

Th2 : \(x+2=1\Rightarrow x=-1\)

Th3 : \(x+2=-3\Rightarrow x=-5\)

Th4 : \(x+3=3\Rightarrow x=0\left(ktm\right)\)

Vậy để P có giá trị nguyên thì x thuộc { - 3 ; - 5 ;- 1 }

\(c,P=-1\Leftrightarrow\frac{3}{x+2}=-1\)

\(\Rightarrow\frac{3}{x+2}=\frac{-1}{1}\Rightarrow3=-1\left(x+2\right)\)

\(\Rightarrow-x-2=3\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

Vậy để P = -1 thì x = - 5

\(d,P>0\Leftrightarrow\frac{3}{x+2}>0\)

Vì \(x+2>0\)nên để \(\frac{3}{x+2}>0\)thì \(x+2>0\)

\(\Rightarrow x>-2\)

Vậy để \(P>0\)thì \(x>2\) và \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

\(đk\hept{\begin{cases}\left(x+2\right)\left(x-2\right)x\ne0\\x+2\ne0\end{cases}< =>x\ne0;x\ne\pm}2\)

P=\(\left(\frac{x}{x^2-4}-\frac{10\left(x-2\right)}{5\left(x+2\right)\left(x-2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right):\)\(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{6-x^2}{x+2}\)

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

b) P \(\in Z\)<=> x-2=3;x-2=-3;x-2=1;x-2=-1 <=> x=5; x=-1; x=3; x=1 (thỏa mãn điều kiện ban đầu)

c) P=1 <=> x-2=3 <=> x=5 (thỏa mãn điều kiện)

d) P>0 <=> x-3 >=0 <=> x>3 kết hợp với điều kiện ban đầu => x>3

b) x^2 - 3x = 0

<=> x(x - 3) = 0

<=> x = 0 hoặc x - 3 = 0

                        x = 3

=> x = 0 hoặc x = 3

Chihiro CTV làm hết đi nhe

\(\left(x-1\right)^2-x\left(5-x\right)=0\)

\(\Leftrightarrow x^2-2x+1-5x+x^2=0\)

\(\Leftrightarrow2x^2-7x+1=0\)

Ta có: \(\Delta=7^2-4.8=25,\sqrt{\Delta}=5\)

\(\orbr{\begin{cases}x_1=\frac{7+5}{4}=3\\x_2=\frac{7-5}{4}=\frac{1}{2}\end{cases}}\)