\(a,\frac{6}{x^2+4x}+\frac{2}{2x+8}=\frac{6}{x\left(x+4\right)}+\frac{2}{2\left(x+4\right)}\)

                                 \(=\frac{6.2}{2x\left(x+4\right)}+\frac{2.x}{2x\left(x+4\right)}=\frac{12}{2x\left(x+4\right)}+\frac{2x}{2x\left(x+4\right)}\)

                                  \(=\frac{12+2x}{2x\left(x+4\right)}=\frac{2\left(6+x\right)}{2x\left(x+4\right)}=\frac{x+6}{x\left(x+4\right)}\)

\(b,\frac{3-2x}{x^2-9}+\frac{1}{2x-6}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}+\frac{1}{2\left(x-3\right)}\)

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

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

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

P/s : Cs sai sót mong thông cảm.

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

\(=\frac{6-4x}{2.\left(x-3\right).\left(x+3\right)}+\frac{x+3}{2.\left(x-3\right).\left(x+3\right)}=\frac{9-3x}{2.\left(x-3\right).\left(x+3\right)}=\frac{-3.\left(x-3\right)}{2.\left(x-3\right).\left(x+3\right)}=-\frac{3}{2.\left(x-3\right)}\)

\(a,ĐKXĐ:\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow x\ne\pm1}\)

\(b,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}+\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)

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

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

      \(=\frac{4x}{1-x^2}\)

\(c,A\ge0\Leftrightarrow\frac{4x}{1-x^2}\ge0\)

               \(\Leftrightarrow\hept{\begin{cases}4x\ge0\\1-x^2\ge0\end{cases}\left(h\right)\hept{\begin{cases}4x\le0\\1-x^2\le0\end{cases}}}\)

              \(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2\le1\end{cases}\left(h\right)\hept{\begin{cases}x\le0\\x^2\ge1\end{cases}}}\)

             \(\Leftrightarrow0\le x\le1\left(h\right)x\le-1\)

Vậy ///////

\(a,\frac{3.\left(x-y\right)}{y-x}=\frac{-3.\left(y-x\right)}{y-z}=-3\)

\(b,\frac{x^2-x}{1-x}=\frac{x.\left(x-1\right)}{1-x}=\frac{-x.\left(1-x\right)}{1-x}=-x\)

\(\frac{3\left(x-y\right)}{y-x}=\frac{3\left(x-y\right)}{-1\left(x-y\right)}=-3\)

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

\(a,2x^2y-xy^2\)

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

\(b,x^4+4x^2-5\)

\(x^4+5x^2-x^2-5\)

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

còn câu b ? :<

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

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

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

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

\(a,\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)

\(b,\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{6}\)

Hok Tốt~~

\(\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)

\(\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{4}\)

Tham khảo nhé~

( x - 1)² + x( x - 4) = 0

x² - 2x + 1 + x² - 4x = 0

2x² - 6x + 1 = 0

2x² - 6x = -1

2x ( x - 3) = -1

      => 2x = -1 => -1/2

hoặc    x - 3 = -1 => x = 2

\(\left(x-1\right)^2+x\left(x-4\right)=0\)

\(x^2-2x+1+x^2-4x=0\)

\(2x^2-6x+1=0\)

\(2.\left[x^2-2.x.1,5+\left(1,5\right)^2\right]-3,5=0\)

\(2.\left(x-1,5\right)^2-3,5=0\)

\(2.\left[\left(x-1,5\right)^2-1,75\right]=0\)

\(\Leftrightarrow\left(x-1,5\right)^2-1,75=0\)

\(\left(x-1,5\right)^2-\left(\sqrt{1,75}\right)^2=0\)

\(\left(x-1,5-\sqrt{1,75}\right)\left(x-1,5+\sqrt{1,75}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1,5-\sqrt{1,75}=0\\x-1,5+\sqrt{1,75}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1,5+\sqrt{1,75}\\x=1,5-\sqrt{1,75}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=1,5+\sqrt{1,75}\\x=1,5-\sqrt{1,75}\end{cases}}\)

\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+80x^3-80x^2+80x+15\)

\(P\left(x\right)=x^7-\left(79+1\right)x^6+\left(79+1\right)x^5-\left(79+1\right)x^4+\left(79+1\right)x^3-\left(79+1\right)x^2+\left(79+1\right)x+15\)

\(P\left(79\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x+15\)

\(P\left(79\right)=x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4-x^3+x^3-x^2+x^2+x+15\)

\(P\left(79\right)=79+15\)

\(P\left(79\right)=94\)

Vậy \(P\left(79\right)=94\)